A. Ancient Diplomacy
建圖，同色點間邊權為$0$，異色點間邊權為$1$，則等價於找一個點使得到它最短路最長的點的最短路最小，Floyd即可。
時間複雜度$O(n^3)$。
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=110,inf=100000000;
int n,m,i,j,k,x,y,a[N],g[N][N];
int main(){
while(~scanf("%d%d",&n,&m)){
if(!n)return 0;
for(i=1;i<=n;i++)scanf("%d",&a[i]);
for(i=1;i<=n;i++)for(j=1;j<=n;j++)g[i][j]=i==j?0:inf;
while(m--){
scanf("%d%d",&x,&y);
g[x][y]=g[y][x]=a[x]^a[y];
}
for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++)g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
int ans=inf;
for(i=1;i<=n;i++){
int now=0;
for(j=1;j<=n;j++)now=max(now,g[i][j]);
ans=min(ans,now);
}
printf("%d\n",ans);
}
}
B. Bob and Banjo
若$AB$線段經過圓$C$的部分長度不超過$t$，則答案為$AB$長度。
否則若$t=0$，則答案為$A$到$C$切線長度$+B$到$C$切線長度$+$一段圓弧。
否則考慮最優路徑，一定是$A$沿直線走到圓上某個點$D$，在圓內長度不超過$t$，然後從$D$開始走若干個長度為$t$的折線到達圓上某個點$E$，再由$E$沿直線到達$B$，中間不經過圓。
列舉順時針還是逆時針走，再列舉$D$到$E$路徑上有多少個$t$，那麼可以解出$DE$這段圓心角的取值範圍，在裡面三分答案即可。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define ms(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e6 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
const double eps = 1e-8, pi = acos(-1.0);
int sgn(double x)
{
if(x < -eps) return -1;
if(x > eps) return 1;
return 0;
}
bool Quadratic(double A, double B, double C, double *t0, double *t1)
{
double discrim = B * B - 4.f * A * C;
if(discrim < 0.) return false;
double rootDiscrim = sqrt(discrim);
double q;
if(B < 0) q = -.5f * (B - rootDiscrim);
else q = -.5f * (B + rootDiscrim);
*t0 = q / A;
*t1 = C / q;
if(*t0 > *t1) swap(*t0, *t1);
return true;
}
struct vec
{
double x, y; vec(){x = y = 0;}
vec(double _x, double _y) {x = _x, y = _y;}
vec operator + (vec v){return vec(x + v.x, y + v.y);}
vec operator - (vec v){return vec(x - v.x, y - v.y);}
vec operator * (double v) {return vec(x * v, y * v);}
vec operator / (double v) {return vec(x / v, y / v);}
double operator * (vec v) {return x * v.x + y * v.y;}
double len(){return hypot(x, y);}
double len_sqr(){return x * x + y * y;}
vec rotate(double c){
return vec(x * cos(c) - y * sin(c), x * sin(c) + y * cos(c));
}
vec trunc(double l){return (*this) * l / len();}
vec tot90(){return vec(-y, x);}
};
vec lerp(vec a, vec b, double t){return a * (1 - t) + b * t;}
struct circle
{
vec c;
double r;
circle(){c = vec(0, 0), r = 0;}
circle(vec _c, double _r){c = _c, r = _r;}
vec point(double a){return vec(c.x + r * cos(a), c.y + r * sin(a));}
};
circle c;
vec st, ed;
double t, ans,th;
int getTangets(vec p, circle C, double *v,double&cen,int mode=0)
{
double x = atan2(p.y - C.c.y, p.x - C.c.x);
double dist = (C.c - p).len();
double ang = acos(C.r / dist);
if(mode){
ang=acos(sqrt(c.r*c.r-t*t/4)/dist);
ang+=th/2;
}
v[0] = x + ang; v[1] = x - ang;
if(v[0]<v[1])swap(v[0],v[1]);
cen=x;
return 2;
}
bool circle_line_intersection(circle c, vec a, vec b, double *t0, double *t1)
{
vec d = b - a;
double A = d * d;
double B = d * (a - c.c) * 2.0;
double C = (a - c.c).len_sqr() - c.r * c.r;
return Quadratic(A, B, C, t0, t1);
}
inline double cal(double x,double offset){
return (st-c.point(x)).len()+(ed-c.point(x+offset)).len();
}
void gao(double v10,double v11,double v20,double v21){
double lim1=v21-v10,lim2=v20-v11;
//if(v10<v11)while(1);
//if(v20<v21)while(1);
//if(v10>v21)while(1);
//if(lim1>lim2)while(1);
lim1/=th;
lim2/=th;
//if(sgn(th)==0)while(1);
int mx=((int)lim2)+10;
int _l=max(1,(int)lim1-10),_r=mx;
//if(lim1>lim2)while(1);
for(int k = max(1,(int)lim1-10); k<=mx; k ++){
//v1[1]<=x<=v1[0]
//v2[1]-th<=x+k*th<=v2[0]
double L=max(v11,v21-k*th);
double R=min(v10,v20-k*th);
if(sgn(L-R)>0)continue;
for(int _=50;_--;){
double len=(R-L)/3;
double m1=L+len,m2=R-len;
double f1=cal(m1,k*th),f2=cal(m2,k*th);
if(f1<f2){
ans=min(ans,f1+k*t);
R=m2;
}else{
ans=min(ans,f2+k*t);
L=m1;
}
//if(L+1e-4>R)break;
}
}
}
int main()
{
while(~ scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &st.x, &st.y, &ed.x, &ed.y, &c.c.x, &c.c.y, &c.r, &t)){
if(sgn(st.x) == 0 && sgn(st.y) == 0 && sgn(ed.x) == 0 && sgn(ed.y) == 0 && sgn(c.c.x) == 0 && sgn(c.c.y) == 0 && sgn(c.r) == 0 && sgn(t) == 0) break;
double t0, t1;
if(circle_line_intersection(c, st, ed, &t0, &t1)){ // 有交點
vec inter1, inter2;
if((sgn(t0)<=0||sgn(t0-1)>=0)&&(sgn(t1)<=0||sgn(t1-1)>=0)){
ans = (st - ed).len();
}else{
inter1 = lerp(st, ed, t0);
inter2 = lerp(st, ed, t1);
if(sgn((inter1 - inter2).len() - t) <= 0){
ans = (st - ed).len();
}
else if(sgn(t)){
ans=1e100;
th = asin(t / 2 / c.r) * 2;
for(int i=0;i<2;i++){
double v1[2],c1, v2[2],c2;
getTangets(st, c, v1,c1,1);
getTangets(ed, c, v2,c2);
for(int i=0;i<10;i++){
v2[0]-=pi*2;
v2[1]-=pi*2;
c2-=pi*2;
}
while(v2[1]<c1)v2[0]+=pi*2,v2[1]+=pi*2,c2+=pi*2;
th = asin(t / 2 / c.r) * 2;
//if(th<0)while(1);
gao(v1[0],v1[1],v2[0],v2[1]);
gao(v1[0],c1,c2,v2[1]);
gao(c1,v1[1],v2[0],c2);
swap(st,ed);
}
//gao(v1[0],c1,c2,v2[1]);
//gao(c1,v1[1],v2[0],c2);
}else{
double v1[2],c1, v2[2],c2;
getTangets(st, c, v1,c1);
getTangets(ed, c, v2,c2);
ans=1e100;
for(int i=0;i<2;i++)for(int j=0;j<2;j++){
vec a=c.point(v1[i]);
vec b=c.point(v2[j]);
double now=(st-a).len()+(ed-b).len();
double o=fabs(v1[i]-v2[j]);
o=min(o,pi*2-o);
ans=min(ans,now+o*c.r);
}
}
}
}
else{
ans = (st - ed).len();
}
if(ans>1e20)while(1);
printf("%.15f\n",ans);
}
return 0;
}
/*
【trick&&吐槽】
0 0 10 0 5 0 3 5
【題意】
【分析】
【時間複雜度&&優化】
*/
C. Chess Knight's Poem
設$v[i][a][b][c][d]$表示匹配了前$i$個字元，兩個棋子分別在$(a,b)$和$(c,d)$是否可能，然後BFS即可。
#include<cstdio>
#include<cstring>
const int N=111;
int dx[8]={-2,-2,2,2,-1,-1,1,1};
int dy[8]={-1,1,-1,1,2,-2,2,-2};
char xiao[4][10]={
{'q','w','e','r','t','y','u','i','o','p'},
{'a','s','d','f','g','h','j','k','l',';'},
{'z','x','c','v','b','n','m',',','.','/'},
{0,0,1,1,1,1,1,1,0,0}
};
char da[4][10]={
{'Q','W','E','R','T','Y','U','I','O','P'},
{'A','S','D','F','G','H','J','K','L',':'},
{'Z','X','C','V','B','N','M','<','>','?'},
{0,0,1,1,1,1,1,1,0,0}
};
int n,i,j,k,x,y,z;
bool v[N][4][10][4][10];
int q[N*40*40][5],h,t;
bool flag;
char a[N];
inline void ext(int o,int A,int B,int C,int D,int _){
if(A<0||A>3)return;
if(B<0||B>9)return;
if(C<0||C>3)return;
if(D<0||D>9)return;
if(A==C&&B==D)return;
if(_==1){
if(xiao[A][B]>1){//not shift not space
if(xiao[C][D]==0){//shift
if(a[o]!=da[A][B])return;
o++;
}else{
if(a[o]!=xiao[A][B])return;
o++;
}
}else if(xiao[A][B]==1){
if(a[o]!=' ')return;
o++;
}
}
if(_==2){
if(xiao[C][D]>1){//not shift not space
if(xiao[A][B]==0){//shift
if(a[o]!=da[C][D])return;
o++;
}else{
if(a[o]!=xiao[C][D])return;
o++;
}
}else if(xiao[C][D]==1){
if(a[o]!=' ')return;
o++;
}
}
if(v[o][A][B][C][D])return;
v[o][A][B][C][D]=1;
q[++t][0]=o;
q[t][1]=A;
q[t][2]=B;
q[t][3]=C;
q[t][4]=D;
}
int main(){
while(1){
gets(a+1);
n=strlen(a+1);
if(a[1]=='*')return 0;
for(i=0;i<=n+5;i++)for(j=0;j<4;j++)for(k=0;k<10;k++)for(x=0;x<4;x++)for(y=0;y<10;y++)v[i][j][k][x][y]=0;
h=1,t=0;
ext(1,3,0,3,9,0);
flag=0;
while(h<=t){
x=q[h][0];
if(x>n){
flag=1;
break;
}
y=q[h][1];
z=q[h][2];
int A=q[h][3];
int B=q[h][4];
h++;
for(int i=0;i<8;i++){
ext(x,y+dx[i],z+dy[i],A,B,1);
ext(x,y,z,A+dx[i],B+dy[i],2);
}
}
puts(flag?"1":"0");
}
}
/*
CAlmimg eventa
*/
D. Drone
將$(t,f(t))$的函式畫圖，可以發現最大值和最小值都在凸殼上，因此答案是可以三分的。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
double x[N], v[N];
double check(double t)
{
double mx = -1e18;
double mn= 1e18;
for(int i = 1; i <= n; ++i)
{
double p = x[i] + t * v[i];
gmax(mx, p);
gmin(mn, p);
}
return mx - mn;
}
int main()
{
while(~scanf("%d", &n), n)
{
for(int i = 1; i <= n; ++i)
{
scanf("%lf%lf", &x[i], &v[i]);
}
double l = 0;
double r = 1.001e10;
for(int tim = 1; tim <= 100; ++tim)
{
double lt = (l + l + r) / 3;
double rt = (l + r + r) / 3;
double lv = check(lt);
double rv = check(rt);
lv <= rv ? r = rt : l = lt;
}
printf("%.12f\n", check(l));
}
return 0;
}
/*
【trick&&吐槽】
2
-100 1
100 -1
3
-100 1
100 -1
101 -1
3
-100 -1
0 0
100 1
0
【題意】
【分析】
【時間複雜度&&優化】
*/
E. Ed and The Legend of Zelda
設$f[i][j][k][l]$表示考慮$i$的子樹，$i$子樹內作弊了$j$次，$i$作弊情況為$k$，$i$有$l$個兒子作弊的方案數，使用組合數轉移。
時間複雜度$O(n^2)$。
#include<cstdio>
const int N=205,P=1000000007;
int n,m,i,x,g[N],nxt[N],f[N][N][2];
int size[N],ans;
int fac[N],inv[N];
int h[N][2][2];
int dp[N][2][2];
inline void up(int&a,int b){a=a+b<P?a+b:a+b-P;}
inline int C(int n,int m){return 1LL*fac[n]*inv[m]%P*inv[n-m]%P;}
void dfs(int x){
size[x]=1;
for(int i=g[x];i;i=nxt[i])dfs(i),size[x]+=size[i];
for(int j=0;j<=1;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)dp[j][k][o]=0;
dp[0][0][0]=fac[size[x]-1];
dp[1][1][0]=fac[size[x]-1];
int pre=1;
for(int i=g[x];i;i=nxt[i]){
for(int j=0;j<=pre+size[i];j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)h[j][k][o]=0;
for(int j=0;j<=pre;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)if(dp[j][k][o]){
for(int A=0;A<=size[i];A++)for(int B=0;B<2;B++)if(f[i][A][B]){
if(k&&B)continue;
if(o&&B)continue;
if(B){
up(h[j+A][k][o+B],1LL*dp[j][k][o]*inv[size[x]-1]%P*fac[size[x]-size[i]-1]%P*C(size[x]-1,size[i]-1)%P*f[i][A][B]%P);
}else{
up(h[j+A][k][o+B],1LL*dp[j][k][o]*inv[size[i]]%P*f[i][A][B]%P);
}
}
}
pre+=size[i];
for(int j=0;j<=pre;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)dp[j][k][o]=h[j][k][o];
}
for(int j=0;j<=size[x];j++)for(int k=0;k<2;k++)f[x][j][k]=0;
for(int j=0;j<=pre;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)
up(f[x][j][k],dp[j][k][o]);
}
int main(){
for(fac[0]=i=1;i<N;i++)fac[i]=1LL*fac[i-1]*i%P;
for(inv[0]=inv[1]=1,i=2;i<N;i++)inv[i]=1LL*(P-inv[P%i])*(P/i)%P;
for(i=2;i<N;i++)inv[i]=1LL*inv[i-1]*inv[i]%P;
while(~scanf("%d%d",&n,&m)){
if(!n)return 0;
for(i=0;i<=n;i++)g[i]=0;
for(i=2;i<=n;i++){
scanf("%d",&x);
nxt[i]=g[x];
g[x]=i;
}
dfs(1);
ans=0;
for(i=0;i<=m;i++)up(ans,f[1][i][0]);
printf("%d\n",ans);
}
}
/*
5 1
1 1 5 1
3 2
1 1
0 0
*/
F. Fix and Solve
首先將每個數轉化為質數出現最多一次的數。
對於每個質數用set維護所有出現的下標，那麼相鄰兩個出現下標可以使得一段區間的非法。
線段樹維護每個區間內被標記為非法的次數最小值以及最小值個數即可。
時間複雜度$O(n\log n\log\log n)$。
#include<cstdio>
#include<set>
#include<algorithm>
#include<vector>
using namespace std;
const int N=100010,M=262150;
int i,j,notprime[N];
vector<int>has[N];
set<int>T[N];
int n,K,m,a[N],all;
int tag[M],mi[M],mcnt[M];
void build(int x,int a,int b){
mi[x]=tag[x]=0;
mcnt[x]=b-a+1;
if(a==b)return;
int mid=(a+b)>>1;
build(x<<1,a,mid);
build(x<<1|1,mid+1,b);
}
inline void tag1(int x,int p){tag[x]+=p;mi[x]+=p;}
void change(int x,int a,int b,int c,int d,int p){
if(c<=a&&b<=d){tag1(x,p);return;}
if(tag[x]){
tag1(x<<1,tag[x]);
tag1(x<<1|1,tag[x]);
tag[x]=0;
}
int mid=(a+b)>>1;
if(c<=mid)change(x<<1,a,mid,c,d,p);
if(d>mid)change(x<<1|1,mid+1,b,c,d,p);
mi[x]=min(mi[x<<1],mi[x<<1|1]);
mcnt[x]=0;
if(mi[x]==mi[x<<1])mcnt[x]+=mcnt[x<<1];
if(mi[x]==mi[x<<1|1])mcnt[x]+=mcnt[x<<1|1];
}
inline int query(){
if(mi[1]>0)return all;
return all-mcnt[1];
}
inline void gao(int l,int r,int p){
if(r-l+1>K)return;
int a=r-K+1,b=l;
change(1,1,all,a,b,p);
}
inline void ins(int x,int y){
T[x].insert(y);
set<int>::iterator it=T[x].find(y),pre=it,nxt=it;
pre--,nxt++;
if(*pre&&*nxt<N)gao(*pre,*nxt,-1);
if(*pre)gao(*pre,y,1);
if(*nxt<N)gao(y,*nxt,1);
}
inline void del(int x,int y){
set<int>::iterator it=T[x].find(y),pre=it,nxt=it;
pre--,nxt++;
if(*pre&&*nxt<N)gao(*pre,*nxt,1);
if(*pre)gao(*pre,y,-1);
if(*nxt<N)gao(y,*nxt,-1);
T[x].erase(y);
}
inline void add(int x,int k){
for(int i=0;i<has[k].size();i++)ins(has[k][i],x);
}
inline void remove(int x,int k){
for(int i=0;i<has[k].size();i++)del(has[k][i],x);
}
int main(){
for(i=2;i<N;i++){
if(!notprime[i]){
for(j=i;j<N;j+=i){
notprime[j]=1;
has[j].push_back(i);
}
}
}
while(~scanf("%d%d%d",&n,&K,&m)){
if(!n)return 0;
all=n-K+1;
build(1,1,all);
for(i=1;i<N;i++)T[i].clear(),T[i].insert(0),T[i].insert(N);
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
add(i,a[i]);
}
printf("%d\n",query());
while(m--){
int x,y;
scanf("%d%d",&x,&y);
remove(x,a[x]);
add(x,a[x]=y);
printf("%d\n",query());
}
long long ans=0;
for(i=1;i<=n;i++)ans+=a[i];
printf("%lld\n",ans);
}
}
G. Gold Bandits
爆搜所有$1$到$2$的最短路，然後再求$1$到$2$的開銷最小的最短路使得收益最大。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 40, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
int d[N], f[N];
int v[N], w[N];
vector<int>a[N];
void bfs()
{
queue<int>q; q.push(1);
MS(d, -1); d[1] = 0;
while(!q.empty())
{
int x = q.front(); q.pop();
for(auto y : a[x])if(d[y] == -1)
{
q.push(y);
d[y] = d[x] + 1;
}
}
}
bool e[N];
int ans;
int sub()
{
MS(e, 0);
MS(f, 63); f[2] = 0;
for(int tim = 1; tim <= n; ++tim)
{
int x = 0;
for(int i = 1; i <= n; ++i)if(!e[i] && f[i] < f[x])
{
x = i;
}
e[x] = 1;
for(auto y : a[x])
{
gmin(f[y], f[x] + w[y]);
}
}
return f[1];
}
void dfs(int x, int val)
{
if(x == 2)
{
gmax(ans, val - sub());
return;
}
for(auto y : a[x])if(d[y] == d[x] + 1)
{
w[y] = v[y];
dfs(y, val + v[y]);
w[y] = 0;
}
}
int main()
{
while(~scanf("%d%d", &n, &m), n || m)
{
for(int i = 1; i <= n; ++i)a[i].clear();
for(int i = 3; i <= n; ++i)scanf("%d", &v[i]);
for(int i = 1; i <= m; ++i)
{
int x, y; scanf("%d%d", &x, &y);
a[x].push_back(y);
a[y].push_back(x);
}
bfs();
ans = 0;
dfs(1, 0);
printf("%d\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
3 3
1
1 2
2 3
1 3
4 4
24 10
1 3
2 3
2 4
1 4
6 8
100 500 300 75
1 3
1 4
3 6
4 5
3 5
4 6
2 5
2 6
7 7
90 1000 700 2000 800
1 3
1 4
1 5
3 7
5 6
2 6
3 6
0 0
【題意】
【分析】
【時間複雜度&&優化】
*/
H. How Many Values?
$O(n\log a)$列舉所有$\gcd$本質不同的區間即可。
#include<cstdio>
const int N=100010;
int n,i,j,l[N],v[N],a[N],ans,vis[N];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main(){
while(~scanf("%d",&n)){
if(!n)return 0;
for(i=0;i<=n;i++)a[i]=l[i]=v[i]=0;
for(ans=i=0;i<=100;i++)vis[i]=0;
for(i=1;i<=n;i++)scanf("%d",&a[i]);
for(i=1;i<=n;i++)for(v[i]=a[i],j=l[i]=i;j;j=l[j]-1){
v[j]=gcd(v[j],a[i]);
while(l[j]>1&&gcd(a[i],v[l[j]-1])==gcd(a[i],v[j]))l[j]=l[l[j]-1];
vis[v[j]]=1;
}
for(i=1;i<=100;i++)if(vis[i])ans++;
printf("%d\n",ans);
}
}
I. Integer Estate Agent
按題意模擬即可。
#include<cstdio>
typedef __int128 lll;
const int N = 1e6 + 10;
int n;
int cnt[N];
void init()
{
int ans = 0;
for(int i = 2; i <= 1e6; i ++){
int x = 0;
for(int j = 0; j <= 1e6; j ++){
x += i + j;
if(x > 1e6) break;
cnt[x] ++;
ans ++;
}
}
//printf("%d\n", ans);
}
int main(){
init();
while(~ scanf("%d", &n), n){
printf("%d\n", cnt[n]);
}
return 0;
}
J. John and Super Mario 169
旅行商問題，狀壓DP即可，時間複雜度$O(2^nn^3)$。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7;
const double inf = 1e18;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
struct P
{
int x, y, z;
}me, sw[13], p[13][13];
double f[1 << 13][13];
double d[13][13];
double dp_sw[1 << 13][13];
double dp[1 << 13][13][13];
int K[13];
double sqr(double x)
{
return x * x;
}
double dis(P a, P b)
{
return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y) + sqr(a.z - b.z));
}
int main()
{
while(~scanf("%d", &n), n)
{
scanf("%d%d%d", &me.x, &me.y, &me.z);
for(int i = 0; i < n; ++i)
{
scanf("%d", &K[i]);
scanf("%d%d%d", &sw[i].x, &sw[i].y, &sw[i].z);
for(int j = 0; j < K[i]; ++j)
{
scanf("%d%d%d", &p[i][j].x, &p[i][j].y, &p[i][j].z);
}
int top = 1 << K[i];
for(int j = 0; j < top; ++j)
{
for(int k = 0; k < K[i]; ++k)f[j][k] = inf;
}
for(int j = 0; j < K[i]; ++j)
{
f[1 << j][j] = dis(sw[i], p[i][j]);
}
for(int sta = 0; sta < top; ++sta)
{
for(int j = 0; j < K[i]; ++j)if(f[sta][j] != inf && (sta >> j & 1))
{
for(int k = 0; k < K[i]; ++k)if(~sta >> k & 1)
{
gmin(f[sta | 1 << k][k], f[sta][j] + dis(p[i][j], p[i][k]));
}
}
}
for(int j = 0; j < K[i]; ++j)
{
d[i][j] = f[top - 1][j];
}
}
int top = 1 << n;
for(int sta = 0; sta < top; ++sta)
{
for(int j = 0; j < n; ++j)
{
dp_sw[sta][j] = inf;
for(int k = 0; k < K[j]; ++k)
{
dp[sta][j][k] = inf;
}
}
}
for(int j = 0; j < n; ++j)
{
dp_sw[1 << j][j] = dis(me, sw[j]);
}
for(int sta = 0; sta < top; ++sta)
{
for(int j = 0; j < n; ++j)
{
if(dp_sw[sta][j] != inf)
{
for(int k = 0; k < K[j]; ++k)
{
gmin(dp[sta][j][k], dp_sw[sta][j] + d[j][k]);
}
}
}
for(int j = 0; j < n; ++j)if(sta >> j & 1)
{
for(int u = 0; u < n; ++u)if(~sta >> u & 1)
{
for(int k = 0; k < K[j]; ++k)if(dp[sta][j][k] != inf)
{
gmin(dp_sw[sta | 1 << u][u], dp[sta][j][k] + dis(p[j][k], sw[u]));
/*
for(int v = 0; v < K[u]; ++v)
{
gmin(dp[sta | 1 << u][u][v],
dp[sta][j][k] + dis(p[j][k], sw[u]) + d[u][v]);
}
*/
}
}
}
}
double ans = inf;
for(int j = 0; j < n; ++j)
{
for(int k = 0; k < K[j]; ++k)
{
gmin(ans, dp[top - 1][j][k]);
}
}
printf("%.10f\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
2 5 5 0
4 6 0 0
7 0 0
-11 -1 0
-11 1 0
-10 0 0
2 5 0 0
0 0 0
0 5 0
0 0 0 0
【題意】
【分析】
【時間複雜度&&優化】
*/