hdu3790 最短路径问题（spfa||dijkstra+两种限制条件）

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http://acm.hdu.edu.cn/showproblem.php?pid=3790

题意：多了一个花费的限制条件，距离相等则选花费少的。

思路：直接在原来的判断条件上加就行，不要想太多。注意有重边。

#include <stdio.h>#include <algorithm>#include <stdlib.h>#include <string.h>#include <iostream>#include <queue>using namespace std;typedef long long LL;const int N = 1005;const int INF = 0x3f3f3f3f;int dis[N], cost[N], G1[N][N], G2[N][N], n;bool vis[N];void dijkstra(int s){    for(int i = 1; i <= n; i++)    {        dis[i] = INF;        cost[i] = INF;    }    dis[s] = 0;    cost[s] = 0;    for(int i = 1; i <= n; i++)    {        int k = -1;        for(int j = 1; j <= n; j++)        {            if(!vis[j] && (k==-1 || dis[j]<dis[k] || (dis[j]==dis[k]&&(cost[j]<cost[k]))))                k = j;        }        if(k == -1) break;        vis[k] = true;        for(int j = 1; j <= n; j++)        {            if(!vis[j] && (dis[k]+G1[k][j]<dis[j] || (dis[k]+G1[k][j]==dis[j]&&(cost[k]+G2[k][j]<cost[j]))))            {                dis[j] = dis[k]+G1[k][j];                cost[j] = cost[k]+G2[k][j];            }        }    }}void spfa(int s){    queue<int>que;    for(int i = 1; i <= n; i++)    {        dis[i] = INF;        cost[i] = INF;    }    dis[s] = 0;    cost[s] = 0;    vis[s] = true;    que.push(s);    while(!que.empty())    {        int now = que.front();        que.pop();        vis[now] = false;        for(int i = 1; i <= n; i++)        {            if(dis[now]+G1[now][i]<dis[i] || (dis[now]+G1[now][i]<dis[i]&&(cost[now]+G2[now][i]<cost[i])))            {                dis[i] = dis[now]+G1[now][i];                cost[i] = cost[now]+G2[now][i];                if(!vis[i])                {                    vis[i] = true;                    que.push(i);                }            }        }    }}int main(){  //  freopen("in.txt", "r", stdin);    int m, u, v, s, t, w1, w2;    while(~scanf("%d%d", &n, &m))    {        if(n == 0 && m == 0) break;        memset(vis, false, sizeof(vis));        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)            {                if(i == j)                {                    G1[i][j] = 0;                    G2[i][j] = 0;                }                else                {                    G1[i][j] = INF;                    G2[i][j] = INF;                }            }        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d%d", &u, &v, &w1, &w2);            if(w1<G1[u][v] || (w1==G1[u][v]&&w2<G2[u][v]))            {                G1[u][v] = G1[v][u] = w1;                G2[u][v] = G2[v][u] = w2;            }        }        scanf("%d%d", &s, &t);      //  dijkstra(s);        spfa(s);        printf("%d %d\n", dis[t], cost[t]);    }    return 0;}

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