题目大意 求多组i到j的最短路径 并输出字典序最小....

现在只会floyd的方式

利用dis[i][j] 表示i到j的路径中i 后面的节点

更新是比较dis[i][j] dis[i][k].

记住这个就好 ，其余存法貌似会有问题。代码如下：

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
const int maxn=50+5;
int N;
int MAP[maxn][maxn];
int dis[maxn][maxn];
int w[maxn];
void input()
{for(int i=1;i<=N;i++)for(int j=1;j<=N;j++)dis[i][j]=1000000000;memset(MAP,0,sizeof(MAP));memset(w,0,sizeof(w));for(int i=1;i<=N;i++)for(int j=1;j<=N;j++){int c;scanf("%d",&c);if(c==-1){MAP[i][j]=1000000000;}else MAP[i][j]=c;}for(int i=1;i<=N;i++){scanf("%d",&w[i]);}
}
int FIND(int a,int b)
{if(dis[a][b]!=1000000000){return FIND(a,dis[a][b]);}else return b;
}
void floyd()
{for(int i=1; i<=N; i++)for(int j=1; j<=N; j++)dis[i][j] = j;for(int k=1;k<=N;k++){for(int i=1;i<=N;i++)for(int j=1;j<=N;j++){int new_patl = MAP[i][k] + MAP[k][j] + w[k];if(MAP[i][j]>new_patl){MAP[i][j]=new_patl;dis[i][j]=dis[i][k];}else  if(MAP[i][j]==new_patl){dis[i][j]=min(dis[i][j],dis[i][k]);}}}
}
void solve()
{int CASE=0;int n,m;int next;while(cin>>n>>m&&(n!=-1&&m!=-1)){printf("From %d to %d :\n",n,m);printf("Path: ");next = n;while(next != m){printf("%d-->", next);next = dis[next][m];}printf("%d\n", next);printf("Total cost : %d\n",MAP[n][m]);printf("\n");}
}
void init()
{freopen("a.in","r",stdin);freopen("a.out","w",stdout);
}
int main()
{// init();while(cin>>N&&N){input();floyd();solve();}
}