hdu 3371 Connect the Cities(最小生成树kruskal)
来源:互联网 发布:用c语言编写一个图形 编辑:程序博客网 时间:2024/06/10 09:20
Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12309 Accepted Submission(s): 3403
Problem Description
In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.
Input
The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
Output
For each case, output the least money you need to take, if it’s impossible, just output -1.
Sample Input
16 4 31 4 22 6 12 3 53 4 332 1 22 1 33 4 5 6
Sample Output
1
Author
dandelion
Source
HDOJ Monthly Contest – 2010.04.04
题目分析:用并查集先将能过在一起的点缩点,然后套最小生成树模板即可
#include <cstring>#include <cstdio>#include <algorithm>#include <iostream>#define MAX 507using namespace std;int n,m,k,t,q;struct Edge{ int u,v,w; bool operator < ( const Edge& a ) const { return w < a.w; }}e[MAX*MAX];int cc = 0;void add ( int u , int v , int w ){ e[cc].u = u; e[cc].v = v; e[cc].w = w; cc++;}int fa[MAX];void init ( ){ for ( int i = 1 ; i <= n ; i++ ) fa[i] =i; cc = 0;}int find ( int x ){ return x==fa[x]?x:fa[x]=find(fa[x]);}void _union ( int x , int y ){ x = find (x); y = find (y); fa[x] = y;}int kruskal ( ){ sort ( e , e+cc ); int sum = 0; for ( int i = 0 ; i < m ; i++ ) { int u = find ( e[i].u ); int v = find ( e[i].v ); if ( u == v ) continue; _union ( u , v ); sum += e[i].w; } int flag = find ( 1 ); for ( int i = 2 ; i <= n ; i ++ ) if ( find(i) != flag ) return -1; return sum;}int main ( ){ int u , v , w; scanf ( "%d" , &t ); while ( t-- ) { scanf ( "%d%d%d" , &n , &m , &k ); init(); for ( int i = 0 ; i < m ; i++ ) { scanf ( "%d%d%d" , &u , &v , &w ); add ( u , v , w ); } for ( int i = 0 ; i < k ; i++ ) { scanf ( "%d" , &q ); scanf ( "%d" , &u ); for ( int j = 1 ; j < q ; j++ ) { scanf ( "%d" , &v ); _union ( u , v ); } } printf ( "%d\n" , kruskal ( ) ); }}
0 0
- hdu 3371 Connect the Cities 最小生成树(kruskal算法)
- hdu 3371 Connect the Cities(最小生成树kruskal)
- HDU 3371 Connect the Cities 并查集+Kruskal算法+最小生成树
- HDU 3371 Connect the Cities 【最小生成树,Prime算法+Kruskal算法】
- 文章标题 HDU 3371 : Connect the Cities(最小生成树--Kruskal+并查集)
- HDU3371 Connect the Cities 【最小生成树Kruskal】
- hdu 3371 Connect the Cities(最小生成树)
- HDU 3371 Connect the Cities(最小生成树)
- hdu 3371 最小生成树Connect the Cities
- hdu oj 3371 Connect the Cities (最小生成树)
- hdu 3371 Connect the Cities(最小生成树))
- HDU 3371 Connect the Cities 最小生成树
- HDU 3371 Connect the Cities 【最小生成树】
- hdu 3371(Connect the Cities)(最小生成树)
- hdu 3371 Connect the Cities 最小生成树
- hdu 3371 Connect the Cities 最小生成树prim
- HDU-3371 Connect the Cities(最小生成树)
- HDU 3371Connect the Cities(最小生成树)
- 在安卓上使用RxJava
- hdu 1272 小希的迷宫
- leetcode_Number of 1 Bits
- java.lang.OutOfMemoryError: PermGen space及其解决方法
- 大巧不工Web前端设计修炼之道——(7)让你的设计更加完美——优化技巧和最佳实践
- hdu 3371 Connect the Cities(最小生成树kruskal)
- ZOJ3623 Battle Ships (完全背包)
- UVA11235
- char varchar varchar2 的区别
- codeforces 294 div2.C
- 蓝桥杯--比酒量
- uvalive 4108(线段树)
- 国企的MySQL选型方案
- java.lang.OutOfMemoryError: Java heap space如何解决?