洛谷P3094 [USACO13DEC]假期计划Vacation Planning

题目描述

有N(1 <= N <= 200)个农场，用1..N编号。航空公司计划在农场间建立航线。对于任意一条航线，选择农场1..K中的农场作为枢纽(1 <= K <= 100, K <= N)。

当前共有M (1 <= M <= 10,000)条单向航线连接这些农场，从农场u_i 到农场 v_i, 将花费 d_i美元。(1 <= d_i <= 1,000,000).

航空公司最近收到Q (1 <= Q <= 10,000)个单向航行请求。第i个航行请求是从农场a_i到农场 b_i，航行必须经过至少一个枢纽农场（可以是起点或者终点农场），因此可能会多次经过某些农场。

请计算可行航行请求的数量，及完成所有可行请求的总费用。

输入输出格式

输入格式：

 

• Line 1: Four integers: N, M, K, and Q.

• Lines 2..1+M: Line i+1 contains u_i, v_i, and d_i for flight i.

• Lines 2+M..1+M+Q: Line 1+M+i describes the ith trip in terms of a_i and b_i

 

输出格式：

 

• Line 1: The number of trips (out of Q) for which a valid route is possible.

• Line 2: The sum, over all trips for which a valid route is possible, of the minimum possible route cost.

 

输入输出样例

输入样例#1：

3 3 1 3 
3 1 10 
1 3 10 
1 2 7 
3 2 
2 3 
1 2 

输出样例#1：

2 
24 

说明

There are three farms (numbered 1..3); farm 1 is a hub. There is a $10 flight from farm 3 to farm 1, and so on. We wish to look for trips from farm 3 to farm 2, from 2->3, and from 1->2.

The trip from 3->2 has only one possible route, of cost 10+7. The trip from 2->3 has no valid route, since there is no flight leaving farm 2. The trip from 1->2 has only one valid route again, of cost 7.

 

 

floyd求出最短路，对于每个询问枚举枢纽找最小花费。

/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int mxn=300;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int mp[mxn][mxn];
int n,m,k,Q;

int main(){
    memset(mp,0x3f,sizeof mp);
    n=read();m=read();k=read();Q=read();
    int i,j;
    int u,v,d;
    for(i=1;i<=m;i++){
        u=read();v=read();d=read();
        mp[u][v]=min(mp[u][v],d);
    }
    for(int e=1;e<=n;e++)
     for(i=1;i<=n;i++)
      for(j=1;j<=n;j++){
          if(i==j)mp[i][j]=0;
          else mp[i][j]=min(mp[i][j],mp[i][e]+mp[e][j]);
      }
    long long ans=0;int cnt=0;
    while(Q--){
        u=read();v=read();
        d=0x3f3f3f3f;
        for(i=1;i<=k;i++){
            d=min(d,mp[u][i]+mp[i][v]);
        }
        if(d<0x3f3f3f3f){
            cnt++;
            ans+=d;
        }
    }
    printf("%d\n%lld\n",cnt,ans);
    return 0;
}