最短路径问题
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 19581 Accepted Submission(s): 5827
Problem Description
给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。
Input
输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)
(1<n<=1000, 0<m<100000, s != t)
Output
输出 一行有两个数， 最短距离及其花费。
Sample Input
3 2 1 2 5 6 2 3 4 5 1 3 0 0
Sample Output
9 11
Source
浙大计算机研究生复试上机考试-2010年
此题比起一般最短路径问题多了一个花费，其他的都没什么，同样不能使用Floyd算法，那这里小编给出SPFA，和Dijkstra两种算法的代码。
SPFA:
#include <iostream>//SPFA
#include <cstdio>
#include <cstdlib>
#include <queue>
#define MAX 0xfffffff
using namespace std;
struct Node
{int p,d;int th;bool operator>(const Node &node)const{if(d == node.d)return p>node.p;return d>node.d;}
};
bool visit[1001];
int mapd[1001][1001],mapp[1001][1001],dis[1001],ps[1001];
priority_queue<Node, vector<Node>, greater<Node> > q;
int main()
{int n,m;int a,b,d,p;while(scanf("%d%d",&n,&m)&&n!=0&&m!=0){for(int i=1; i<=n; i++){visit[i] = false;dis[i] = MAX;ps[i] = MAX;for(int j=1;j<=n;j++){if(i == j){mapd[i][j] = mapp[i][j] = 0;continue;}mapd[i][j] = MAX;mapp[i][j] = MAX;}}for(int i=0; i<m; i++){scanf("%d%d%d%d",&a,&b,&d,&p);if(d<mapd[a][b]){mapd[a][b] = mapd[b][a] = d;mapp[a][b] = mapp[b][a] = p;}if(d == mapd[a][b] && p < mapp[a][b]){mapp[a][b] = mapp[b][a] = p;}}int s,t;scanf("%d%d",&s,&t);Node node;node.th = s;node.d = 0;node.p = 0;q.push(node);while(!q.empty()){Node node = q.top();q.pop();if(visit[node.th]) continue;visit[node.th] = true;for(int i=1; i<=n; i++){if(i!=node.th && mapd[node.th][i] != MAX && visit[i] == false){if(node.d + mapd[node.th][i] < dis[i]){dis[i] = node.d + mapd[node.th][i];ps[i] = node.p + mapp[node.th][i];Node tmp;tmp.th = i;tmp.d = dis[i];tmp.p = ps[i];q.push(tmp);}else if(node.d + mapd[node.th][i] == dis[i] && node.p + mapp[node.th][i] < ps[i]){ps[i] = node.p + mapp[node.th][i];Node tmp;tmp.th = i;tmp.d = dis[i];tmp.p = ps[i];q.push(tmp);}}}}if(s == t)printf("%d %d\n",0,0);else printf("%d %d\n",dis[t],ps[t]);}return 0;
}
#include <stdio.h>//Dijkstra算法
#include <iostream>
#include <cstring>
#define INF 0xfffffff
#define N 1000+10
using namespace std;
int map[N][N],cost[N][N];
int min_dis,min_cost;
void Dijkstra(int s,int t,int n)
{int dis[N],cost1[N];bool vis[N];memset(vis,false,sizeof(vis));for(int i=1; i<N; i++){dis[i] = map[s][i];cost1[i] = cost[s][i];}dis[s] = 0;vis[s] = true;for(int i=1; i<n; i++){int min = INF;int k;for(int j=1; j<=n; j++){if(!vis[j] && min>dis[j]){min = dis[j];k = j;}}vis[k] = true;for(int j=1; j<=n; j++){if(!vis[j] && dis[j]>min+map[k][j]){dis[j] = min+map[k][j];cost1[j] = cost1[k]+cost[k][j];}else if(!vis[j] && dis[j] == min + map[k][j] && cost1[j] > cost[k][j] + cost1[k]){cost1[j] = cost[k][j] + cost1[k];}}if(vis[t]){min_dis = dis[t];min_cost = cost1[t];return;}}min_dis = dis[t];min_cost = cost1[t];
}
int main()
{int n,m;while(scanf("%d%d",&n,&m)&&(n||m)){for(int i=1; i<=n; i++)for(int j=1; j<i; j++){map[i][j] = map[j][i] = INF;cost[i][j] = cost[j][i] = INF;}while(m--){int a,b,d,p;scanf("%d%d%d%d",&a,&b,&d,&p);if(map[a][b] > d){map[a][b] = map[b][a] = d;cost[a][b] = cost[b][a] = p;}}int s,t;scanf("%d%d",&s,&t);Dijkstra(s,t,n);printf("%d %d\n",min_dis,min_cost);}return 0;
}