hdu1596——find the safest road

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find the safest road

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7275    Accepted Submission(s): 2580


Problem Description
XX星球有很多城市,每个城市之间有一条或多条飞行通道,但是并不是所有的路都是很安全的,每一条路有一个安全系数s,s是在 0 和 1 间的实数(包括0,1),一条从u 到 v 的通道P 的安全度为Safe(P) = s(e1)*s(e2)…*s(ek) e1,e2,ek是P 上的边 ,现在8600 想出去旅游,面对这这么多的路,他想找一条最安全的路。但是8600 的数学不好,想请你帮忙 ^_^
 

Input
输入包括多个测试实例,每个实例包括:
第一行:n。n表示城市的个数n<=1000;
接着是一个n*n的矩阵表示两个城市之间的安全系数,(0可以理解为那两个城市之间没有直接的通道)
接着是Q个8600要旅游的路线,每行有两个数字,表示8600所在的城市和要去的城市
 

Output
如果86无法达到他的目的地,输出"What a pity!",
其他的输出这两个城市之间的最安全道路的安全系数,保留三位小数。
 

Sample Input
31 0.5 0.50.5 1 0.40.5 0.4 131 22 31 3
 

Sample Output
0.5000.4000.500
 

Author
#include <set>#include <map>#include <list>#include <stack>#include <queue>#include <vector>#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 1010;struct node{double weight;int next;int to;}edge[N * N];double dist[N];int head[N];int tot, n;void addedge(int from, int to, double weight){edge[tot].weight = weight;edge[tot].to = to;edge[tot].next = head[from];head[from] = tot++;}void spfa(int v0){for (int i = 1; i <= n; i++){dist[i] = 0.000;}dist[v0] = 1.000;queue <int> qu;while ( !qu.empty() ){qu.pop();}qu.push(v0);while ( !qu.empty() ){int u = qu.front();qu.pop();for (int i = head[u]; ~i; i = edge[i].next){int v = edge[i].to;//printf("%.3f %.3f\n", dist[v], dist[u] * edge[i].weight);if (dist[v] < dist[u] * edge[i].weight){dist[v] = dist[u] * edge[i].weight;qu.push(v);}}}}int main(){while (~scanf("%d", &n)){double w;int u, v;memset (head, -1, sizeof(head) );tot = 0;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){scanf("%lf", &w);addedge(i, j, w);}}int m;scanf("%d", &m);while (m--){scanf("%d%d", &u, &v);spfa(u);if (dist[v] != 0){printf("%.3lf\n", dist[v]);continue;}printf("What a pity!\n");}}return 0;}

ailyanlu
 

Source
HDU 2007-Spring Programming Contest - Warm Up (1)
 

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最短路变形, 只要把限制条件改一下就可以了
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