【程序設計大賽刷題記錄】C語言 02

【Tallest cow】
鏈接：https://ac.nowcoder.com/acm/contest/999/C
來源：牛客網

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.

【代碼】
#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;
第一個是鍵的類型，第二個是值的類型。
【總結】
運用遞歸算法來求出第i頭牛的最大可能高度，本題的難點就在於理解輸入和輸出，特別是輸入的後面幾行。