hdu 3790 最短路径问题 (spfa模板题)
题意:给出n个点m条无向边,每条边有一个距离和一个花费。给出s,t。问从s到t的最短距离以及最短距离时的最小花费。当有多个距离最短的方案时,选取花费最少的。
先写几道题加深理解。
记得初始化。。。。。。
/* ***********************************************
Author :111qqz
Created Time :2016年05月21日 星期六 18时42分24秒
File Name :code/hdu/3790.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1E3+7;
7int n,m;
8int d[N];
9int p[N];
10bool inq[N];
11vector < pair<int,pair<int,int> > > edge[N];
int s,t;
1void init()
2{
3 ms(inq,false);
4 ms(d,0x3f);
5 ms(p,0x3f);
for ( int i = 0 ; i <=n ; i++) edge[i].clear();
}
1void spfa()
2{
3 queue<int>q;
4 q.push(s);
5 d[s] = 0 ;
6 p[s] = 0;
7 inq[s]=true;
1 while (!q.empty())
2 {
3 int now = q.front();
4// cout<<"now:"<<now<<endl;
5 q.pop();
6 inq[now] = false;
1 int siz = edge[now].size();
2 for ( int i = 0 ; i < siz ; i ++)
3 {
4 int v = edge[now][i].fst;
5 int nd = edge[now][i].sec.fst;
6 int nw = edge[now][i].sec.sec;
7 // cout<<"v:"<<v<<" nd:"<<nd<<" nw:"<<nw<<endl;
8 if (d[v]>d[now]+nd||((d[v]==d[now]+nd)&&(p[v]>p[now]+nw)))
9 {
10 d[v] = d[now] + nd;
11 p[v] = p[now] + nw;
1 if (inq[v]) continue;
2 q.push(v);
3 inq[v] = true;
4 }
5 }
6 }
1}
2int main()
3{
4 #ifndef ONLINE_JUDGE
5 freopen("code/in.txt","r",stdin);
6 #endif
1 while (scanf("%d%d",&n,&m)!=EOF)
2 {
3 if (n==0&&m==0) break;
4 init();
5 for ( int i = 1 ; i <= m ; i++)
6 {
7 int u,v,d,w;
8 scanf("%d%d%d%d",&u,&v,&d,&w);
9 edge[u].push_back(MP(v,MP(d,w)));
10 edge[v].push_back(MP(u,MP(d,w)));
11 }
scanf("%d%d",&s,&t);
spfa();
printf("%d %d\n",d[t],p[t]);
}
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}