POJ 2528 Mayor's posters （線段樹 區間更新+離散化）

Mr_Treeeee發表於2020-04-06
原文網址 : https://blog.csdn.net/mr_treeeee/article/details/75807619
Mayor's posters
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 65645 Accepted: 18967

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules: 
  • Every candidate can place exactly one poster on the wall. 
  • All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown). 
  • The wall is divided into segments and the width of each segment is one byte. 
  • Each poster must completely cover a contiguous number of wall segments. 

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers li and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= li <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered li, li+1 ,... , ri. 

Output

For each input data set print the number of visible posters after all the posters are placed. 

The picture below illustrates the case of the sample input. 

Sample Input

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

題意：
給你無限長的廣告牌，給你n個廣告和他們放置的位置，按照輸入資料的順序放置前後，問你能看見幾個廣告，看見一部分也算。

POINT：

2018.6.1:

從後往前處理。

線段樹只存有沒有空位。

然後有空位就可以change l到r。都讓他沒空位就行了。

**每個數不是代表一個邊界而是代表一塊展板！**

所以去年理解錯了題目



#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 10000*2+33;

#define lt x<<1
#define rt x<<1|1


int num[maxn<<3];
int lazy[maxn<<3];
int x[maxn*4];

int sz;

int findi(int xx)
{
	return lower_bound(x+1,x+1+sz,xx)-(x);
}


struct node
{
	int l,r,s;
}b[10100];

void up(int x)
{
	num[x]=num[lt]|num[rt];
}


void build(int x,int l,int r)
{
	if(l==r) num[x]=1;
	else{
		int mid = (l+r)>>1;
		build(lt,l,mid);
		build(rt,mid+1,r);
		up(x);
	}
	lazy[x]=0;
}

void pushdown(int x)
{
	if(lazy[x]){
		num[lt]=0;
		num[rt]=0;
		lazy[lt]=lazy[rt]=1;
		lazy[x]=0;
	}
}

void change(int x,int l,int r,int ll,int rr)
{
	if(ll<=l&&rr>=r){
		num[x]=0;
		lazy[x]=1;
		return;
	}
	pushdown(x);
	int mid = (l+r)>>1;
	if(ll<=mid) change(lt,l,mid,ll,rr);
	if(mid<rr) change(rt,mid+1,r,ll,rr);
	up(x);
}

int query(int x,int l,int r,int ll,int rr)
{
	if(ll<=l&&rr>=r){
		if(num[x]>0)
			return 1;
		return 0;
	}
	pushdown(x);
	int mid = (l+r)>>1;
	int ans=0;
	if(ll<=mid){
		ans=query(lt,l,mid,ll,rr);
	}
	if(ans==0&&mid<rr){
		ans=query(rt,mid+1,r,ll,rr);
	}
	up(x);
	return ans;
}
int flag[10200];

int main()
{
	//每個數不是代表一個邊界而是代表一塊展板！
	int n;
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		memset(num,0,sizeof num);
		int cnt=0;
		x[++cnt]=0;
		x[++cnt]=10000000;
		for(int i=1;i<=n;i++){
			scanf("%d%d",&b[i].l,&b[i].r);
			x[++cnt]=b[i].l;
			x[++cnt]=b[i].l+1;
			x[++cnt]=max(1,b[i].l-1);
			x[++cnt]=b[i].r;
			x[++cnt]=b[i].r+1;
			x[++cnt]=max(1,b[i].r-1);
		}
		sort(x+1,x+1+cnt);
		sz=unique(x+1,x+1+cnt)-(x+1);
		int R=findi(10000000);

		build(1,1,R);

		int ans=0;
		for(int i=n;i>=1;i--){
			int l=findi(b[i].l);
			int r=findi(b[i].r);
			if(query(1,1,R,l,r)){
				ans++;
				change(1,1,R,l,r);
			}
		}
		printf("%d\n",ans);


	}


}















歷史：
先離散化，但是如果直接按照順序去離散化，理論上是錯的，但是可以AC.
如果你AC了 這組資料必然過不了
1
3
1 10
1 3
6 10

正確是3 但是隻當你算出來是2的時候你才能A
為什麼呢
因為4-5這段區間離散化的時候並沒有處理到
區間儲存的是被第幾個廣告完全覆蓋，有多個廣告則是零，這題折磨了我很久。

更悲催的是明明已經AC了，我卻看成WA又檢查了20分鐘！！！

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <map>
#include <string.h>
#include <algorithm>
#include <fstream>
using namespace std;
#define lt 2*x
#define rt 2*x+1
#define LL long long
const int N  = 22000*5;
int b[N/5];
int col[N],fl[N];
struct node
{
    int l,r;
}a[N/5];
void pushdown(int x)
{
    fl[lt]=fl[x];
    fl[rt]=fl[x];
    col[lt]=fl[x];
    col[rt]=fl[x];
    fl[x]=0;
    
}
void add(int x,int l,int r,int ll,int rr,int i)
{
    if(fl[x]) pushdown(x);
    if(ll<=l&&rr>=r)
    {
        col[x]=i;
        fl[x]=i;
    }
    else
    {
        int mid=(l+r)>>1;
        if(ll<=mid) add(lt,l,mid,ll,rr,i);
        if(mid<rr) add(rt,mid+1,r,ll,rr,i);
        if(col[lt]==col[rt]) col[x]=col[rt];
        else col[x]=0;
    }
}
int vis[N];
int query(int x,int l,int r)
{
    if(fl[x]) pushdown(x);
    int ans=0;
    if(col[x]||l==r)
    {
        if(!vis[col[x]])
        {
            vis[col[x]]=1;
            ans++;
        }
    }
    else
    {
        int mid=(l+r)>>1;
        ans+=query(lt,l,mid);
        ans+=query(rt,mid+1,r);
    }
    return ans;
    
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        memset(b,0,sizeof b);
        memset(col,0,sizeof col);
        memset(fl,0,sizeof fl);
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&a[i].l,&a[i].r);
            b[++cnt]=a[i].l;
            b[++cnt]=a[i].r;
        }
        sort(b+1,b+1+cnt);
        int m=unique(b+1,b+1+cnt)-b-1;
        for(int i=1;i<=n;i++)
        {
            a[i].l=lower_bound(b+1,b+m+1,a[i].l)-b;
            a[i].r=lower_bound(b+1,b+m+1,a[i].r)-b;
        }
        for(int i=1;i<=n;i++)
        {
            add(1,1,m,a[i].l,a[i].r,i);
        }
        memset(vis,0,sizeof vis);
        int ans=query(1,1,m);
        printf("%d\n",ans);
    }
    return 0;
}


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