【程序设计大赛刷题记录】C语言 02

【Tallest cow】
链接：https://ac.nowcoder.com/acm/contest/999/C
来源：牛客网

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.

【代码】
#include<bits/stdc++.h>
using namespace std;
const int N = 10100;
int s[N];
map<pair<int, int>, bool> m;
int main() {
int n, u, v, r, x, y;
cin >> n >> u >> v >> r;
s[1] = v;
for(int i=1; i<=r; i++) {
cin >> x >> y;
if(x > y) swap(x, y);
if(!m[make_pair(x, y)]) {
s[x+1] -= 1, s[y] += 1;
m[make_pair(x, y)] = 1;
}
}
int ans = 0;
for(int i=1; i<=n; i++){
cout << s[i]+ans << endl;
ans += s[i];
}
return 0;
}

【map的用法】
map<type1name,type2name>maps;
第一个是键的类型，第二个是值的类型。
【总结】
运用递归算法来求出第i头牛的最大可能高度，本题的难点就在于理解输入和输出，特别是输入的后面几行。