需要处理好字典序最小的路径
HDU1385(ZOJ1456)-Minimum Transport
//Hdu1385-ZOJ1456
//给定邻接矩阵,求给定起点到终点的最短路径,若有相同路长的路径按照字典序输出
//Floyd比较适合此题
//网上看到的两种做法比较推荐
//第一种是Floyd+记录起点后继
//第二种是Floyd+深搜(从起点开始深搜至终点)-利用Floyd得到的最短路剪枝
//其他的解法则是按照SPFA,dijkstra解,当路长相同时需要进行比较(较繁琐)
//Time:0Ms Memory:404K
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std; #define INF 0x3f3f3f3f
#define MAX 105 int n;
int d[MAX][MAX];
int board[MAX][MAX], c[MAX];
int path[MAX][MAX]; //记录i到j最短路径中i的后继 void floyd()
{
memcpy(d, board, sizeof(d));
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] + c[k])
{
d[i][j] = d[i][k] + d[k][j] + c[k];
path[i][j] = path[i][k];
}
//相同路径下选择后继更小的
else if (d[i][j] == d[i][k] + d[k][j] + c[k] && path[i][j] > path[i][k])
path[i][j] = path[i][k];
}
} void output(int s,int e)
{
printf("-->%d", path[s][e]);
if (path[s][e] != e)
output(path[s][e], e);
} int main()
{
while (scanf("%d", &n), n)
{
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
scanf("%d", &board[i][j]);
if (board[i][j] == -1) board[i][j] = INF;
path[i][j] = j;
}
for (int i = 1; i <= n; i++)
scanf("%d", &c[i]);
floyd();
int s, e;
while (scanf("%d%d", &s, &e), s != -1 && e != -1)
{
printf("From %d to %d :\n", s, e);
printf("Path: %d", s);
if(s != e) output(s, e); //起点与终点不同开始递归
printf("\nTotal cost : %d\n\n", d[s][e]);
}
}
return 0;
}