如果借助顶点1,任意两点之间的最短路径更新为:
借助顶点1的方法是:
#include <iostream>
using namespace std;
int main() {
int e[11][11],n,m,p,q,t;
int inf = 99999999;//正无穷的定义,即使两个正无穷相加,结果也不超过int类型的范围
cin >> n >> m;
for(int i = 1; i <=n; i++) {
for(int j = 1; j <= n; j++) {
if(i == j) {
e[i][j] = 0;
} else {
e[i][j] = inf;
}
}
}
for(int i = 1; i <=m; i++) {
cin>>p>>q>>t;
e[p][q] = t;
}
for(int k = 1; k <=n; k++) {
for(int i = 1; i <=n; i++) {
for(int j = 1; j <= n; j++) {
if(e[i][j] > e[i][k] + e[k][j]) {
e[i][j] = e[i][k] + e[k][j];
}
}
}
}
for(int i = 1; i <=n; i++) {
for(int j = 1; j <= n; j++) {
cout << e[i][j] << " ";
}
cout << endl;
}
return 0;
}