hdu 1385 求字典序最小的最短路

需要注意的是这个题不仅有边权还有点权，起点和终点的点权不算。

思路还是一样，只是把路过的点的点权也加上再松弛即可。另外，距离相等的时候需要判断一下，选择字典序小的链连到这个点，方法就是把这个点以及之前的点放到栈里比较。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <stack>
#include <cstdio>
using namespace std;

const int INF = 100000001;
const int N = 10001;
int graph[N][N];
int dist[N];
int tax[N];
int pre[N];
bool visit[N];
int n, m;

bool lessThan( int cur, int x, int y )
{
    stack<int> sx, sy;
    //注意：cur必须也放到栈里
    sx.push(cur);
    sy.push(cur);
    while ( x != -1 )
    {
        sx.push(x);
        x = pre[x];
    }
    while ( y != -1 )
    {
        sy.push(y);
        y = pre[y];
    }
    while ( !sx.empty() && !sy.empty() )
    {
        if ( sx.top() != sy.top() ) return sx.top() < sy.top();
        sx.pop();
        sy.pop();
    }
    return sx.size() < sy.size();
}

void dij( int s, int t )
{
    memset( visit, false, sizeof(visit) );
    for ( int i = 0; i <= n; i++ )
    {
        dist[i] = INF;
    }
    dist[s] = 0;
    pre[s] = -1;
    for ( int i = 1; i <= n; i++ )
    {
        int u = 0;
        for ( int j = 1; j <= n; j++ )
        {
            if ( !visit[j] && dist[j] < dist[u] )
            {
                u = j;
            }
        }
        visit[u] = true;
        for ( int j = 1; j <= n; j++ )
        {
            if ( visit[j] ) continue;
            int d = dist[u] + graph[u][j] + tax[j];
            if ( d < dist[j] )
            {
                dist[j] = d;
                pre[j] = u;
            }
            else if ( d == dist[j] && lessThan( j, u, pre[j] ) )
            {
                pre[j] = u;
            }
        }
    }
}

void out( int i )
{
    if ( pre[i] == -1 )
    {
        printf("Path: %d", i);
    }
    else
    {
        out( pre[i] );
        printf("-->%d", i);
    }
}

int main ()
{
    while ( scanf("%d", &n), n )
    {
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= n; j++ )
            {
                scanf("%d", &graph[i][j]);
                if ( graph[i][j] == -1 )
                {
                    graph[i][j] = INF;
                }
            }
        }
        for ( int i = 1; i <= n; i++ )
        {
            scanf("%d", tax + i);
        }
        int s, t;
        while ( scanf("%d%d", &s, &t) != EOF )
        {
            if ( s == -1 || t == -1 ) break;
            int tmp = tax[t];
            tax[t] = 0;
            dij( s, t );
            tax[t] = tmp;
            printf("From %d to %d :
", s, t);
            out( t );
            printf("
Total cost : %d

", dist[t]);
        }
    }
    return 0;
}