uva 10273 - Eat or Not to Eat?(暴力枚举)
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题目大意:uva 10273 - Eat or Not to Eat?
题目大意:在一个农场里有n头牛,每头牛都有一个产奶周期,然后n行就是每头牛的产奶周期;然后农场主想要杀一些奶牛来吃,所以每天就把产奶量最小的杀来吃;如果有两只产奶量一样小,那今天就不杀。输出没有杀掉的牛的个数,以及最后杀牛的日子。
解题思路:暴力枚举, 注意说停止的时间,即当前天数 - 上一次吃牛的日子 > tmp(最小周期);然后如果没有吃过牛,最晚杀牛的日子是0.
#include <stdio.h>#include <string.h>const int N = 1005;const int M = 15;const int INF = 1 << 30;const int tmp = 2520;int n, s[N], g[N][M], vis[N];void init() {scanf("%d\n", &n);for (int i = 0; i < n; i++) {scanf("%d", &s[i]);for (int j = 0; j < s[i]; j++)scanf("%d", &g[i][j]);}}bool find(int d, int& id) {int Min = INF;bool flag = false;for (int i = 0; i < n; i++) {if (vis[i]) continue;int c = d % s[i];if (Min > g[i][c]) {id = i;Min = g[i][c];flag = true;} else if (Min == g[i][c])flag = false;}return flag;}void solve() {int id, del = -1, cnt = 0;memset(vis, 0, sizeof(vis));for (int i = 0; i - del <= tmp; i++) {if (find(i, id)) {del = i;vis[id] = 1;cnt++;}}printf("%d %d\n", n - cnt, del + 1);}int main () {int cas;scanf("%d", &cas);while (cas--) {init();solve();}return 0;}
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