(阶段三 dijkstra1.4)HDU 1596 find the safest road(最短路的变形题:求乘积,求最大值)

#include <iostream>#include <cstdio>using namespace std;const int maxn = 1100;const int inf = -100;double map[maxn][maxn];double d[maxn];int s[maxn];int n, m;int start, end;double dijkstra(int v) {int i;for (i = 1; i <= n; ++i) {s[i] = 0;d[i] = map[v][i];}s[v] = 1;d[v] = 1;int j;for (i = 1; i < n; ++i) {double min = inf;int pos;for (j = 1; j <= n; ++j) {if (!s[j] && min < d[j]) {pos = j;min = d[j];}}s[pos] = 1;/** * 此题是最短路径的变形体。 * 原来最短路的特征是: 求最短路,求最小值 * 而这道题是求乘积,求最大值 */for (j = 1; j <= n; ++j) {if (!s[j] && (d[j] < (d[pos] * map[pos][j]))) {d[j] = d[pos] * map[pos][j];}}}return d[end]; //返回所要求的源节点到n节点的最短路径}int main() {while (scanf("%d", &n) != EOF) {int i, j;for (i = 1; i <= n; ++i) { //初始化..所有的节点之间都不相通for (j = 1; j <= n; ++j) {map[i][j] = inf;}}double temp;for(i = 1 ; i <= n ; ++i){for(j = 1 ; j <= n ; ++j){scanf("%lf",&temp);map[i][j] = temp;}}int q;scanf("%d",&q);while(q--){scanf("%d%d",&start,&end);if(start == end){printf("%.3lf\n",map[start][end]);}else {if(dijkstra(start) > 0){printf("%.3lf\n",dijkstra(start));}else{printf("What a pity!\n");}}}}return 0;}