Silver Cow Party POJ 3268(最短路)

原题

题目链接

题目分析

了解题意后会发现跑两次最短路就能解决问题了,首先考虑奶牛们回家时的路径,其实只要从聚餐点跑一次最短路,记为dis2[],而考虑奶牛们出发时的最短路,只需要把所有边反置一下,再从聚餐点跑一遍最短路,记为dis[],最后只需要找出牛i使得dis[i]+dis2[i]的值最小即为答案.普通的最短路可以用dijkstra跑,复杂度更低.

代码

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <utility>
  4 #include <cstdio>
  5 #include <cmath>
  6 #include <cstring>
  7 #include <string>
  8 #include <vector>
  9 #include <stack>
 10 #include <queue>
 11 #include <map>
 12 #include <set>
 13 
 14 using namespace std;
 15 typedef long long LL;
 16 const int INF_INT=0x3f3f3f3f;
 17 const LL INF_LL=0x3f3f3f3f3f3f3f3f;
 18 
 19 typedef pair<int,int> P;
 20 int edge[2000][2000];
 21 int dis[2000],redis[2000];
 22 
 23 int n,m,x;
 24 
 25 void dijkstra1(int s)
 26 {
 27     for(int i=1;i<=n;i++) dis[i]=INF_INT;
 28     priority_queue<P,vector<P>,greater<P> > que;
 29     que.push(P(0,s));
 30     dis[s]=0;
 31     while(que.size())
 32     {
 33         P p=que.top();que.pop();
 34         int from=p.second;
 35         if(p.first>dis[from]) continue;
 36         for(int i=1;i<=n;i++)
 37         {
 38             if(edge[from][i]!=INF_INT)
 39             {
 40                 if(dis[i]>dis[from]+edge[from][i])
 41                 {
 42                     dis[i]=dis[from]+edge[from][i];
 43                     que.push(P(dis[i],i));
 44                 }
 45             }
 46         }
 47     }
 48 }
 49 
 50 void dijkstra2(int s)
 51 {
 52     for(int i=1;i<=n;i++) redis[i]=INF_INT;
 53     priority_queue<P,vector<P>,greater<P> > que;
 54     que.push(P(0,s));
 55     redis[s]=0;
 56     while(que.size())
 57     {
 58         P p=que.top();que.pop();
 59         int from=p.second;
 60         if(p.first>redis[from]) continue;
 61         for(int i=1;i<=n;i++)
 62         {
 63             if(edge[from][i]!=INF_INT)
 64             {
 65                 if(redis[i]>redis[from]+edge[from][i])
 66                 {
 67                     redis[i]=redis[from]+edge[from][i];
 68                     que.push(P(redis[i],i));
 69                 }
 70             }
 71         }
 72     }
 73 }
 74 
 75 int main()
 76 {
 77 //    freopen("black.in","r",stdin);
 78 //    freopen("black.out","w",stdout);
 79     cin>>n>>m>>x;
 80     for(int i=1;i<=n;i++)
 81         for(int j=1;j<=n;j++) edge[i][j]=INF_INT;
 82     for(int i=0;i<m;i++)
 83     {
 84         int from,to,cost;
 85         scanf("%d %d %d",&from,&to,&cost);
 86         edge[from][to]=min(edge[from][to],cost);
 87     }
 88  /*   for(int i=1;i<=n;i++)
 89         for(int j=1;j<=n;j++) if(edge[i][j]!=INF_INT) printf("edge %d %d %d\n",i,j,edge[i][j]);*/
 90     dijkstra1(x);
 91 //    for(int i=1;i<=n;i++) printf("dis[%d]=%d\n",i,dis[i]);
 92     for(int i=1;i<=n;i++)
 93         for(int j=1;j<i;j++) swap(edge[i][j],edge[j][i]);
 94   /*  for(int i=1;i<=n;i++)
 95         for(int j=1;j<=n;j++) if(edge[i][j]!=INF_INT) printf("edge %d %d %d\n",i,j,edge[i][j]);*/
 96     dijkstra2(x);
 97  //   for(int i=1;i<=n;i++) printf("redis[%d]=%d\n",i,redis[i]);
 98     int ans=0;
 99     for(int i=1;i<=n;i++) ans=max(ans,dis[i]+redis[i]);
100     cout<<ans<<endl;
101     return 0;
102 }

 

posted @ 2019-08-27 15:15  VBL  阅读(105)  评论(0)  编辑  收藏  举报