hdu 1385 Minimum Transport Cost 打印路径 + 路径字典序最小

//除了边的权值之外每个点还附带一个权值，所以在松弛操作的时候要把点的权值也计算进去
//另外在总费用最小的情况下要输出字典序最小的路径，同样是在松弛操作那里处理
//如果能更新d[i]使d[i]变小则直接更新

//如果是与d[i]相同则判断一下如果更新的话会不会使路径的字典序更小，如果能才更新否则不更新；这一点和记录花费那很像

#include <iostream>
#include <memory.h>
#include <stdio.h>
using namespace std;
#define Max_V  1005
const int IN = (1<<28);
int G[105][105];
int Path[105][105];
int Tax[105];
int N,S,D;
void Floyd()
{
    for( int u = 1; u <= N; u++ )
        for( int v = 1; v <= N; v++ )
            for( int w = 1; w <= N; w++ )
            {
                if( G[v][w] > G[v][u] + G[u][w] + Tax[u] )
                {
                    G[v][w] = G[v][u] + G[u][w]+ Tax[u];
                    Path[v][w] = Path[v][u];
                }
                else if( G[v][w] == G[v][u] + G[u][w] + Tax[u] )
                {
                    if( Path[v][w] > Path[v][u] )
                    {
                        Path[v][w] = Path[v][u];
                    }
                }
            }
}
void Print(int i, int j)
{
    if( i==j )
    {
        printf("%d",i);
        return;
    }
    printf("%d-->",i);
    Print(Path[i][j],j);
}
int main()
{
    while( cin >> N, N!=0)
    {
        memset(G, 0, sizeof(G));
        memset(Path, 0, sizeof(Path));
        for( int i = 1; i <= N; i++ )
        {
            for( int j = 1; j <= N; j++ )
            {
                Path[i][j] = j;
                cin >> G[i][j];
                if( G[i][j] == -1 )
                {
                    G[i][j] = IN;
                }
            }
        }
        for( int i = 1; i <= N; i++ )
            cin >> Tax[i];
        Floyd();
        while( cin >> S >> D, S != -1 || D != -1 )
        {
            if( S==D )
            {
                printf("From %d to %d :\n",S,D);
                printf("Path: %d\n",S);
                printf("Total cost : %d\n\n", 0);
                continue;
            }
            printf("From %d to %d :\n",S,D);
            printf("Path: ");
            Print(S,D);
            printf("\n");
            printf("Total cost : %d\n\n", G[S][D]);
        }
    }

    return 0;
}