[USACO07JAN]区间统计Tallest Cow

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.

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.

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.

给出牛的可能最高身高，然后输入m组数据 a b,代表a，b可以相望，最后求所有牛的可能最高身高输出

输入输出格式

输入格式：

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

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

输出格式：

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

输入输出样例

输入样例#1： 复制

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

输出样例#1： 复制

5
4
5
3
4
4
5
5
5

有头牛从１到n线性排列，每头牛的高度为h[i]现在告诉你这里面的牛的最大高度为maxH,而且有r组关系，每组关系输入两个数字，假设为a和b,表示第a头牛能看到第b头牛，能看到的条件是a, b之间的其它牛的高度都严格小于min(h[a], h[b]),而h[b] >= h[a]
最后求所有牛的可能最高身高输出

先将身高全部假设为最高身高
把给的信息(a,b)去重之后，为了使[a+1,b-1]内所有数都小于v[a],v[b]只需要区间减1即可
维护一个差分数组，在x[a]++,x[b+1]--然后扫一遍就可以得到每个位置的值

//2018年2月14日17:41:54
#include <iostream>
#include <cstdio>
#include <map>
using namespace std;

const int N = 100001;

int n, id, f[N], m;
map<pair<int, int>, bool> mp;

int main(){
    scanf("%d%d%d%d", &n, &id, &f[1], &m);
    while(m--){
        int x, y;
        scanf("%d%d", &x, &y);
        if(x > y) swap(x, y);
        if(x != y){
            if(mp[pair<int, int>(x, y)] != 1){
                mp[pair<int, int>(x, y)] = 1;
                f[x+1]--;
                f[y]++;
            }
        }
    }
    for(int i=1, now=0;i<=n;i++)
        printf("%d
", now+=f[i]);

    return 0;
}