标签：acm   图论   王桂平   

题意  给你一个无向图的邻接矩阵  和途径每个点需要的额外花费首尾没有额外花费  求图中某两点之间的最短路并打印字典序最小路径

要求多组点之间的就用floyd咯  打印路径也比较方便  nex[i][j]表示从i点到j点最短路的第一个途经点  那么如果路径中加入一个节点k后 nex[i][j]应该更新为nex[i][k]  因为要途径k了

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 100,INF=0x3f3f3f;
int cost[N][N], tax[N], nex[N][N];
int s, t, n;

void floyd()
{
    int tmp;
    for(int k = 1; k <= n; ++k)
    for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= n; ++j)
    {
        tmp = cost[i][k] + cost[k][j] + tax[k];
        if(cost[i][j] > tmp || (cost[i][j] == tmp && nex[i][j] > nex[i][k]))
        {
            nex[i][j] = nex[i][k];
            cost[i][j] = tmp;
        }
    }
}

int main()
{
    while(scanf("%d", &n), n)
    {
        memset(nex,0,sizeof(nex));
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= n; ++j)
            {
                scanf("%d", &cost[i][j]);
                if(cost[i][j] < 0) cost[i][j]=INF;
                else nex[i][j]=j;
            }
        }

        for(int i = 1; i <= n; ++i)  scanf("%d", &tax[i]);
        floyd();
        while(scanf("%d%d", &s, &t), s > 0)
        {
            int k=s;
            printf("From %d to %d :\nPath: %d", s, t, s);
            while(k!=t) printf("-->%d", k=nex[k][t]);
            printf("\nTotal cost : %d\n\n", cost[s][t]);
        }
    }
    return 0;
}

Minimum Transport Cost

5
0 3 22 -1 4
3 0 5 -1 -1
22 5 0 9 20
-1 -1 9 0 4
4 -1 20 4 0
5 17 8 3 1
1 3
3 5
2 4
-1 -1
0

From 1 to 3 :
Path: 1-->5-->4-->3
Total cost : 21

From 3 to 5 :
Path: 3-->4-->5
Total cost : 16

From 2 to 4 :
Path: 2-->1-->5-->4
Total cost : 17

HDU 1385 Minimum Transport Cost (字典序打印最短路)

标签：acm   图论   王桂平   

原文地址：http://blog.csdn.net/acvay/article/details/40616925