hdu1385 最短路 floyd + dfs

Minimum Transport Cost

题意很简单，有N个城市，两两之间的距离用一个邻接矩阵给出。问题关键在于，找出最短路，并输出字典序最小的一条。以前一直也没想过这个问题，没想过怎么把最短的一条输出来。其实也不较简单，就是先用Floyd求得点对距离，在按照字典序dfs，把最短距离作为一个强剪枝来用，得到的第一个合法路径即为最短最小字典序路径。路径的输出可以用递归的形式来实现。

 

#include<stdio.h>
#include<string.h>
#define INF 0xfffffff
#define NN 100
int N, S, T, ok;
int dis[NN][NN];
int map[NN][NN];
int mark[NN];
int pre[NN];
int cost[NN];
void floyd(){
     int i, j, k;
     memcpy(dis, map, sizeof(map));
     for (k = 1; k <= N; k++){
         for (i = 1; i <= N; i++){
             for (j = 1; j <= N; j++){
                 if (dis[i][j] > dis[i][k] + dis[k][j] + cost[k]){
                    dis[i][j] = dis[i][k] + dis[k][j] + cost[k];
                 }
             }
         }
     }
}

void dfs(int cur, int d){
     int i, t;
     if (d > dis[S][T]) return;
     if (cur == T && d == dis[S][T]){
        ok = 1;
        return;
     }
     for (i = 1; i <= N; i++){
         if (!mark[i]){
            mark[i] = 1;
            pre[i] = cur;
            t = d + map[cur][i];
            if (i != T){
               t += cost[i];
            }
            dfs(i, t);
            if (ok) return;
            mark[i] = 0;
         }
     }
}

void out(int cur){
     if (cur == S){
        printf("%d", cur);
        return;
     }
     out(pre[cur]);
     printf("-->%d", cur);
}
int main()
{
    int i, j, a, b;
    //freopen("q1456.in", "r", stdin);
    //freopen("a.out", "w", stdout);
    while(scanf("%d", &N) != EOF){
        if (N == 0) break;
        for (i = 1; i <= N; i++){
            for (j = 1; j <= N; j++){
                scanf("%d", &a);
                if (a == -1) a = INF;
                map[i][j] = a;
            }
        }
        for (i = 1; i <= N; i++){
            scanf("%d", &cost[i]);
        }
        floyd();
        while(scanf("%d%d", &a, &b) != EOF){
            if (a == - 1 && b == -1) break;
            S = a;
            T = b;
            ok = 0;
            memset(mark, 0, sizeof(mark));
            dfs(S, 0);
            printf("From %d to %d :\nPath: ", a, b);
            out(b);puts("");
            printf("Total cost : %d\n", dis[a][b]);
            puts("");
        }
    }
    return 0;
}