HDU - 1385 Minimum Transport Cost（floyd打印路径）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1385
题意：给你一张图的邻接矩阵和每个城市的收费，最后给你若干个起点和终点，求最短路径并打印
思路：多源最短路想到的肯定是floyd，用path[i][j]来表示从i到j经过的第一个点是哪个。
代码：#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;

int n;
int d[maxn][maxn],tax[maxn],path[maxn][maxn];
void floyd(){
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
path[i][j]=j;
}
}
for (int k=1;k<=n;k++){
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
if (d[i][j]>d[i][k]+d[k][j]+tax[k]){
d[i][j]=d[i][k]+d[k][j]+tax[k];
path[i][j]=path[i][k];
}
else if (d[i][j]==d[i][k]+d[k][j]+tax[k]){
path[i][j]=min(path[i][j],path[i][k]);
}
}
}
}
}
int main () {
while (~scanf ("%d",&n)&&n){
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
int cost;
scanf ("%d",&cost);
d[i][j]=cost;
if (cost==-1) d[i][j]=INF;
}
}
for (int i=1;i<=n;i++) scanf ("%d",&tax[i]);
int s,t;
floyd();
while (~scanf ("%d%d",&s,&t)){
if (s==-1&&t==-1) break;
printf ("From %d to %d :\nPath: ",s,t);
int to=s;
while (to!=t){
printf ("%d-->",to);
to=path[to][t];
}
printf ("%d\n",t);
printf ("Total cost : %d\n\n",d[s][t]);
}
}
return 0;
}