[2016-04-02][POJ][3268][Silver Cow Party]

• 时间:2016-04-02 20:30:40 星期六

• 题目编号:[2016-04-02][POJ][3268][Silver Cow Party]

• 题目大意:给定一个有向图图,1~n牛到k牛出参加party,所有牛都走最快的路,问所有牛中往返,耗时最长为多少,

• 分析:

  • 求从1 -> k的最短路和 k -> 1的最短路之和,
  • 从1跑一次Dijkstra,从k跑一次Dijkstra,求和,
    • 在discuss发现有个叫矩阵转置的东西,即k->x的最短路,相当于把所有边反转之后,x -> k的最短路

#include <queue>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 1000 + 10;
struct Node{
        int v,c;
        Node(int _v = 0,int _c = 0):v(_v),c(_c){}
        bool operator < (const Node & a)const {
                return c > a.c;
        }
};
int vis[maxn],d1[maxn],d2[maxn],a[maxn][maxn],n,m,x;
void Dijkstra(int d[],int s){
        priority_queue< Node > q;
        memset(vis,0,sizeof(vis));
        memset(d,0x3f,sizeof(int) * (n + 1));
        d[s] = 0;
        q.push(Node(s,0));
        Node tmp;
        while(!q.empty()){
                tmp = q.top();q.pop();
                int u = tmp.v;
                if(vis[u])      continue;
                vis[u] = 1;
                for(int i = 1;i <= n ; ++i){
                        if(!vis[i]&& a[u][i] && d[i] > d[u] + a[u][i]){
                                d[i] = d[u] + a[u][i];
                                q.push(Node(i,d[i]));
                        }
                }
        }
}
void trans(){
        for(int i = 1;i <= n ;++i)
                for(int j = i + 1;j <= n ; ++j)
                        swap(a[i][j],a[j][i]);       
}
int main(){
        int cntcase = 0,_a,b,c;
        int t;
        scanf("%d%d%d",&n,&m,&x);
        memset(a,0,sizeof(a));
        for(int i = 0;i < m ; ++i){
                scanf("%d%d%d",&_a,&b,&c);
                a[_a][b] = c;
        }
        Dijkstra(d1,x);
        trans();
        Dijkstra(d2,x);
        int ans = 0;
        for(int i = 1;i <= n ; ++i){
                ans = max(ans,d2[i]+d1[i]);
        }
        printf("%d\n\n",ans);
        return 0;
}

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