# 洛谷 P2879 [USACO07JAN]区间统计Tallest Cow

2021年09月15日 阅读数：3

## 题目描述

FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow.ios

FJ has made a list of R (0 ≤ R ≤ 10,000) lines of the form "cow 17 sees cow 34". This means that cow 34 is at least as tall as cow 17, and that every cow between 17 and 34 has a height that is strictly smaller than that of cow 17.ide

For each cow from 1..N, determine its maximum possible height, such that all of the information given is still correct. It is guaranteed that it is possible to satisfy all the constraints.spa

## 输入输出格式

Line 1: Four space-separated integers: N, I, H and Rget

Lines 2..R+1: Two distinct space-separated integers A and B (1 ≤ A, B ≤ N), indicating that cow A can see cow B.string

Lines 1..N: Line i contains the maximum possible height of cow i.it

## 输入输出样例

```9 3 5 5
1 3
5 3
4 3
3 7
9 8```

```5
4
5
3
4
4
5
5
5思路：贪心。首先所有赋值为最大值，而后进行贪心。去重后，把区间l，r之间的值均减去一。或者先判断区间l，r之间的最大值，若是比两个端点都就不进行修改，防止修改。不去重```
```#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 10010
using namespace std;
int n,m,pos,maxh;
int h[MAXN];
int main(){
scanf("%d%d%d%d",&n,&pos,&maxh,&m);
for(int i=1;i<=n;i++)    h[i]=maxh;
for(int i=1;i<=m;i++){
int l,r;
scanf("%d%d",&l,&r);
if(l>r)    swap(l,r);
if(abs(l-r)!=1){
int maxn=-1;
for(int j=l+1;j<r;j++)
maxn=max(maxn,h[j]);
if(maxn>=min(h[l],h[r]))
for(int j=l+1;j<r;j++)
h[j]-=1;
}
}
for(int i=1;i<=n;i++)
cout<<h[i]<<endl;
}```
```去重#include<set>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 10010
using namespace std;
set<int>se[MAXN];
int n,m,pos,maxh;
int h[MAXN];
int main(){
scanf("%d%d%d%d",&n,&pos,&maxh,&m);
for(int i=1;i<=n;i++)    h[i]=maxh;
for(int i=1;i<=m;i++){
int l,r;
scanf("%d%d",&l,&r);
if(l>r)    swap(l,r);
if(se[l].find(r)!=se[l].end())    continue;
se[l].insert(r);
if(abs(l-r)!=1)
for(int j=l+1;j<r;j++)
h[j]-=1;
}
for(int i=1;i<=n;i++)
cout<<h[i]<<endl;
}```

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