《算法笔记》10.4小节——图算法专题-＞最短路径 问题 E: 最短路径问题

题目描述

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。

输入

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t。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)

输出

输出 一行有两个数， 最短距离及其花费。

样例输入

3 2
1 2 5 6
2 3 4 5
1 3
0 0

样例输出

9 11

分析： dijkstra 模板题，和 pta 的 1003 做法几乎一致。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0x3fffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;typedef struct node
{int dist,cost;
}node;node graph[1001][1001];void dijkstra(int n,int s,int t,int d[],int price[])
{bool vis[n+5]={0};d[s]=0;price[s]=0;for(int times=0;times<n;++times){int u=-1,mini=INF;for(int i=0;i<=n;++i){if(!vis[i]&&d[i]<mini)u=i,mini=d[i];}if(u==-1)return;vis[u]=1;for(int i=0;i<=n;++i){if(!vis[i]&&graph[u][i].dist!=INF){if(d[i]>d[u]+graph[u][i].dist){d[i]=d[u]+graph[u][i].dist;price[i]=price[u]+graph[u][i].cost;}else if(d[i]==d[u]+graph[u][i].dist){if(price[i]>price[u]+graph[u][i].cost)price[i]=price[u]+graph[u][i].cost;}}}}
}int main(void)
{#ifdef testfreopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);clock_t start=clock();#endif //testint n,m,s,t;while(scanf("%d%d",&n,&m),n){int d[n+5]={0},price[n+5]={0};int pre[n+5];for(int i=0;i<=n;++i){d[i]=INF;for(int j=0;j<=n;++j)graph[i][j].dist=INF;}for(int i=0;i<m;++i){int a,b,c,z;scanf("%d%d%d%d",&a,&b,&c,&z);graph[a][b].dist=c,graph[a][b].cost=z;graph[b][a]=graph[a][b];}scanf("%d%d",&s,&t);dijkstra(n,s,t,d,price);printf("%d %d\n",d[t],price[t]);}#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位#endif //testreturn 0;
}