``````#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;

}``````