POJ3263 Tallest Cow【差分数组】

Tallest Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 5486 Accepted: 2524

Description

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.

Input

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.

Output

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

Sample Input
9 3 5 5
1 3
5 3
4 3
3 7
9 8

Sample Output
5
4
5
3
4
4
5
5
5

Source
USACO 2007 January Silver

问题链接：POJ3263 Tallest Cow
问题简述：（略）
问题分析：
输入n、i、h和r，总共n头牛，最高高度为h，给定r个关系。这r个关系由a和b描述，第a头牛和第b头牛互相看得见（有坑，不一定满足a<=b的关系）。问这n头牛的高度。
假设这n头牛的基础高度为h，a和b互相看得到，需要把a和b之间牛的高度减去1。用差分实现比较容易，然后算一下前缀和。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ3263 Tallest Cow */#include <iostream>
#include <algorithm>
#include <set>
#include <cstdio>
#include <cstring>using namespace std;const int N = 10000;
int d[N + 1];int main()
{int n, i, h, r, a, b;set<pair<int, int> > s;scanf("%d%d%d%d", &n, &i, &h, &r);memset(d, 0, sizeof(d));while(r--) {scanf("%d%d", &a, &b);if(a > b) swap(a, b);if(s.find(make_pair(a, b)) != s.end()) continue;else s.insert(make_pair(a, b));d[a + 1]--;d[b]++;}for(int i = 1; i <= n; i++)printf("%d\n", h + (d[i] += d[i - 1]));return 0;
}

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