HDU 3080 The plan of city rebuild（prim和kruskal）

果7發表於2014-04-29
Rebuild

The plan of city rebuild

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 616    Accepted Submission(s): 215


Problem Description
News comes!~City W will be rebuilt with the expectation to become a center city. There are some villages and roads in the city now, however. In order to make the city better, some new villages should be built and some old ones should be destroyed. Then the officers have to make a new plan, now you , as the designer, have the task to judge if the plan is practical, which means there are roads(direct or indirect) between every two villages(of course the village has not be destroyed), if the plan is available, please output the minimum cost, or output"what a pity!".
 

Input
Input contains an integer T in the first line, which means there are T cases, and then T lines follow.
Each case contains three parts. The first part contains two integers l(0<l<100), e1, representing the original number of villages and roads between villages(the range of village is from 0 to l-1), then follows e1 lines, each line contains three integers a, b, c (0<=a, b<l, 0<=c<=1000), a, b indicating the village numbers and c indicating the road cost of village a and village b . The second part first contains an integer n(0<n<100), e2, representing the number of new villages and roads(the range of village is from l to l+n-1), then follows e2 lines, each line contains three integers x, y, z (0<=x, y<l+n, 0<=z<=1000), x, y indicating the village numbers and z indicating the road cost of village x and village y. The third part contains an integer m(0<m<l+n), representing the number of deserted villages, next line comes m integers, p1,p2,…,pm,(0<=p1,p2,…,pm<l+n) indicating the village number. 
Pay attention: if one village is deserted, the roads connected are deserted, too.
 

Output
For each test case, If all villages can connect with each other(direct or indirect), output the minimum cost, or output "what a pity!".
 

Sample Input
2
4 5
0 1 10
0 2 20
2 3 40
1 3 10
1 2 70
1 1
4 1 60
2
2 3
3 3
0 1 20
2 1 40
2 0 70
2 3
0 3 10
1 4 90
2 4 100
0





 



Sample Output




70
160





 



Author



wangjing



 








題目大意：先給你n1個點，m1條邊，又給你n2個點，m2條邊，然後告訴你有些點不需要，然後求剩下點的最





小生成樹。反正光看題目就花了半小時，真不明白了，老路，新路，給邊還分兩次給。。





用了prim和kruskal兩種演算法來實現，中間還debug了蠻久。












Prim演算法：


#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int INF =1e8;
int mp[205][205];
int des[205];
int visi[205];
int low[205];  //更新最小值
int n;

int prim()
{
    int i,j;

    memset(visi,0,sizeof(visi));
    for(i=0;i<=200;i++)
        low[i]=INF;
    int ans=0,pos=-1;
    for(i=0;i<n;i++)
    {
        if(!des[i])
        {
            pos=i;
            break;
        }
    }

    if(pos==-1) return 0;

    for(i=0;i<=200;i++)
        low[i]=mp[pos][i];

    visi[pos]=1;
    int mi;
    for(i=0;i<n;i++)
    {
        mi=INF;
        int flag=0;
        for(j=0;j<n;j++)
        {
            if(des[j]) continue;
            if(!visi[j])
            {
                flag=1;
                if(low[j]<mi)
                {
                    mi=low[j];
                    pos=j;
                }
            }
        }

        if(mi==INF&&flag)
            return -1;
        if(!flag) return ans;
        ans+=mi;
        visi[pos]=1;
        for(j=0;j<n;j++)
            if(!visi[j])
                low[j]=min(low[j],mp[pos][j]);
    }
    return ans;
}

int main()
{
    int tes,i,j,res;
    scanf("%d",&tes);

    while(tes--)
    {
        int n1,m,u,v,val;
        scanf("%d%d",&n1,&m);

        for(i=0;i<=200;i++)
            for(j=0;j<=200;j++)
                mp[i][j]=INF;
        memset(des,0,sizeof(des));
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&val);
            if(mp[u][v]>val)   //這個地方，邊可能會多次給出，坑。。
            {
                mp[u][v]=val;
                mp[v][u]=val;
            }
        }
        int n2;
        scanf("%d%d",&n2,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&val);
            if(mp[u][v]>val)
            {
                mp[u][v]=val;
                mp[v][u]=val;
            }
        }
        n=n1+n2;
        int t,x;
        scanf("%d",&t);
        for(i=0;i<t;i++)
        {
            scanf("%d",&x);
            des[x]=1;
        }

        res=prim();
        if(res==-1) puts("what a pity!");
        else printf("%d\n",res);
    }
    return 0;
}





Kruskal演算法：


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF =1e6;
const int maxn=205;

int n,m,t;   //總點數，總邊數，desert的邊數

int des[maxn];  //為1代表城市荒廢
int fa[maxn];

struct Edge
{
    int from;
    int to;
    int val;
}edge[maxn*maxn];

int cmp(Edge p1,Edge p2)
{
    //if(des[p1.from]||des[p1.to]) return 0;
    //if(des[p2.from]||des[p2.to]) return 1;
    return p1.val<p2.val;
}

void init()
{
    for(int i=0;i<=200;i++)
        fa[i]=i;
}

int find1(int x)
{
    if(fa[x]!=x) fa[x]=find1(fa[x]);
    return fa[x];
}

void merge1(int p1,int p2)
{
    p1=find1(p1);
    p2=find1(p2);
    fa[p1]=p2;
}

int Kruskal()
{
    sort(edge,edge+m,cmp);
    int ans=0,ste=0;

    for(int i=0;i<m&&ste<n-t-1;i++)
    {
        int u=edge[i].from,v=edge[i].to,val=edge[i].val;
        if(des[u]||des[v]) continue;
        if(find1(u)!=find1(v))
        {
            ste++;
            ans+=val;
            merge1(u,v);
        }
    }

    if(ste==n-t-1) return ans;
    return -1;
}

int main()
{
    int tes,i,j,res;
    scanf("%d",&tes);

    while(tes--)
    {
        int n1,m1,u,v,val,n2,m2;
        scanf("%d%d",&n1,&m1);

        memset(des,0,sizeof(des));

        for(i=0;i<m1;i++)
        {
            scanf("%d%d%d",&u,&v,&val);
            edge[i].from=u;
            edge[i].to=v;
            edge[i].val=val;
        }

        scanf("%d%d",&n2,&m2);
        n=n1+n2;
        m=m1+m2;

        for(i=m1;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&val);
            edge[i].from=u;
            edge[i].to=v;
            edge[i].val=val;
        }

        int x;
        scanf("%d",&t);
        for(i=0;i<t;i++)
        {
            scanf("%d",&x);
            des[x]=1;
        }

        init();
        res=Kruskal();

        //cout<<n<<" :n"<<endl;
        if(res==-1) puts("what a pity!");
        else printf("%d\n",res);
    }
    return 0;
}









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