HDU6187-Destroy Walls

Destroy Walls

                                                                         Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
                                                                                                   Total Submission(s): 272    Accepted Submission(s): 117

Problem Description

Long times ago, there are beautiful historic walls in the city. These walls divide the city into many parts of area. 

Since it was not convenient, the new king wants to destroy some of these walls, so he can arrive anywhere from his castle. We assume that his castle locates at (0.6∗2√,0.6∗3√). 

There are n towers in the city, which numbered from 1 to n. The ith's location is (xi,yi). Also, there are m walls connecting the towers. Specifically, the ith wall connects the tower ui and the tower vi(including the endpoint). The cost of destroying the ith wall is wi. 

Now the king asks you to help him to divide the city. Firstly, the king wants to destroy as less walls as possible, and in addition, he wants to make the cost least. 

The walls only intersect at the endpoint. It is guaranteed that no walls connects the same tower and no 2 walls connects the same pair of towers. Thait is to say, the given graph formed by the walls and towers doesn't contain any multiple edges or self-loops.

Initially, you should tell the king how many walls he should destroy at least to achieve his goal, and the minimal cost under this condition.

 

Input

There are several test cases.

For each test case:

The first line contains 2 integer n, m.

Then next n lines describe the coordinates of the points.

Each line contains 2 integers xi,yi. 

Then m lines follow, the ith line contains 3 integers ui,vi,wi

|xi|,|yi|≤105

3≤n≤100000,1≤m≤200000

1≤ui,vi≤n,ui≠vi,0≤wi≤10000

 

Output

For each test case outout one line with 2 integers sperate by a space, indicate how many walls the king should destroy at least to achieve his goal, and the minimal cost under this condition. 

 

Sample Input

4 4-1 -1-1 11 11 -11 2 12 3 23 4 14 1 2

 

Sample Output

1 1

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

Recommend

liuyiding

 

题意：有一个n个结点m条边的带边权无向平面图，有一个人在一个区域，要拆一些墙使得他可以到达任意一个区域，问最小花费

解题思路：生成一棵最大生成树即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cctype>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const int maxn = 1e5 + 10;int n, m, x, y,f[100009];struct node{int u, v;LL w;bool operator<(const node &a)const{return w > a.w;}}a[200009];int Find(int x){return f[x] == x ? x : f[x] = Find(f[x]);}int main(){while (~scanf("%d%d", &n, &m)){LL ans = 0;for (int i = 1; i <= n; i++) scanf("%d%d", &x, &y);for (int i = 1; i <= m; i++) scanf("%d%d%lld", &a[i].u, &a[i].v, &a[i].w),ans+=a[i].w;sort(a + 1, a + 1 + m);for (int i = 1; i <= n; i++) f[i] = i;int cnt = 0;for (int i = 1; i <= m; i++){int uu = Find(a[i].u), vv = Find(a[i].v);if (uu == vv) { cnt++; continue; }f[uu] = vv;ans -= a[i].w;}printf("%d %lld\n", cnt, ans);}return 0;}