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。

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=300;
 9 int read(){
10     int x=0,f=1;char ch=getchar();
11     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
12     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
13     return x*f;
14 }
15 int mp[mxn][mxn];
16 int n,m,k,Q;
17
18 int main(){
19     memset(mp,0x3f,sizeof mp);
20     n=read();m=read();k=read();Q=read();
21     int i,j;
22     int u,v,d;
23     for(i=1;i<=m;i++){
24         u=read();v=read();d=read();
25         mp[u][v]=min(mp[u][v],d);
26     }
27     for(int e=1;e<=n;e++)
28      for(i=1;i<=n;i++)
29       for(j=1;j<=n;j++){
30           if(i==j)mp[i][j]=0;
31           else mp[i][j]=min(mp[i][j],mp[i][e]+mp[e][j]);
32       }
33     long long ans=0;int cnt=0;
34     while(Q--){
35         u=read();v=read();
36         d=0x3f3f3f3f;
37         for(i=1;i<=k;i++){
38             d=min(d,mp[u][i]+mp[i][v]);
39         }
40         if(d<0x3f3f3f3f){
41             cnt++;
42             ans+=d;
43         }
44     }
45     printf("%d\n%lld\n",cnt,ans);
46     return 0;
47 }