UVA - 10099 The Tourist Guide kruskal算法

来源：互联网发布：手机游戏充值端口代理编辑：程序博客网时间：2024/05/22 10:57

题目大意：导游要带人去旅游，需要乘坐公交车，公交车有人数限制，导游要把所有人都带到终点，问导游最少要带几趟才能将所有人带到终点

解题思路：和噪音的那题相似，要从起点带到终点的次数最少，起点到终点的最小值就要达到最大，小心一点的就是，每次带过去的人应该要减去1个导游

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define INF 0x3f3f3f3f#define maxn 105int N,R;int len[maxn][maxn];int main(){int mark = 1;while(scanf("%d%d",&N,&R) != EOF && N + R) {int t1,t2,l;for(int i = 1; i <= N; i++)for(int j = 1; j <= N; j++)if(i == j)len[i][j] = 0;elselen[i][j] = -1;for(int i = 0; i < R; i++) {scanf("%d%d%d",&t1,&t2,&l);len[t1][t2] = l;len[t2][t1] = l;}scanf("%d%d%d",&t1,&t2,&l);for(int k = 1; k <= N; k++)for(int i = 1; i <= N; i++)for(int j = 1; j <= N; j++)if(len[i][k] != -1 && len[k][j] != -1) {int temp = min(len[i][k],len[k][j]);if(len[i][j] == -1 || temp > len[i][j])len[i][j] = temp;}int ans;if(l % (len[t1][t2] - 1) == 0)ans = l / (len[t1][t2] - 1);elseans = l / (len[t1][t2] - 1) + 1;printf("Scenario #%d\n",mark++);printf("Minimum Number of Trips = %d\n\n",ans);}return 0;}

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