首页 > 代码库 > 并查集+贪心 HDU4424
并查集+贪心 HDU4424
先对边进行从大到小的排序
并查集保存节点数和长度
Conquer a New Region
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1289 Accepted Submission(s): 429
Problem Description
The wheel of the history rolling forward, our king conquered a new region in a distant continent.
There are N towns (numbered from 1 to N) in this region connected by several roads. It‘s confirmed that there is exact one route between any two towns. Traffic is important while controlled colonies are far away from the local country. We define the capacity C(i, j) of a road indicating it is allowed to transport at most C(i, j) goods between town i and town j if there is a road between them. And for a route between i and j, we define a value S(i, j) indicating the maximum traffic capacity between i and j which is equal to the minimum capacity of the roads on the route.
Our king wants to select a center town to restore his war-resources in which the total traffic capacities from the center to the other N - 1 towns is maximized. Now, you, the best programmer in the kingdom, should help our king to select this center.
There are N towns (numbered from 1 to N) in this region connected by several roads. It‘s confirmed that there is exact one route between any two towns. Traffic is important while controlled colonies are far away from the local country. We define the capacity C(i, j) of a road indicating it is allowed to transport at most C(i, j) goods between town i and town j if there is a road between them. And for a route between i and j, we define a value S(i, j) indicating the maximum traffic capacity between i and j which is equal to the minimum capacity of the roads on the route.
Our king wants to select a center town to restore his war-resources in which the total traffic capacities from the center to the other N - 1 towns is maximized. Now, you, the best programmer in the kingdom, should help our king to select this center.
Input
There are multiple test cases.
The first line of each case contains an integer N. (1 <= N <= 200,000)
The next N - 1 lines each contains three integers a, b, c indicating there is a road between town a and town b whose capacity is c. (1 <= a, b <= N, 1 <= c <= 100,000)
The first line of each case contains an integer N. (1 <= N <= 200,000)
The next N - 1 lines each contains three integers a, b, c indicating there is a road between town a and town b whose capacity is c. (1 <= a, b <= N, 1 <= c <= 100,000)
Output
For each test case, output an integer indicating the total traffic capacity of the chosen center town.
Sample Input
4 1 2 2 2 4 1 2 3 1 4 1 2 1 2 4 1 2 3 1
Sample Output
4 3
Source
2012 Asia ChangChun Regional Contest
#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <limits.h> #include <ctype.h> #include <string.h> #include <string> #include <algorithm> #include <iostream> #include <queue> #include <stack> #include <deque> #include <vector> #include <set> #include <map> using namespace std; #define MAXN 200000 + 200 typedef __int64 LL; struct node{ LL x; LL y; LL cost; }E[MAXN]; LL f[MAXN];//父亲节点 LL dis[MAXN];//与该节点相连所有的点到该节点的权值的和 LL sum[MAXN];//该节点的儿子数 bool cmp(node a,node b){ return a.cost > b.cost; } LL find(LL x){ if(x == f[x]){ return x; } return f[x] = find(f[x]); } void init(){ LL i; for(i=0;i<MAXN;i++){ f[i] = i; dis[i] = 0; sum[i] = 1; } } void unio(LL x,LL y,LL v){ f[x] = y; sum[y]+=sum[x]; dis[y] = v; } int main(){ LL n,i; while(~scanf("%I64d",&n)){ for(i=1;i<n;i++){ scanf("%I64d%I64d%I64d",&E[i].x,&E[i].y,&E[i].cost); } sort(E+1,E+n,cmp); init(); for(i=1;i<n;i++){ LL dx = find(E[i].x); LL dy = find(E[i].y); LL dis1 = dis[dx]+E[i].cost*sum[dy];//因为E【i】.cost一定要比这两个集合中的任何一条边都要短 LL dis2 = dis[dy]+E[i].cost*sum[dx]; if(dis1 > dis2){ unio(dy,dx,dis1); } else{ unio(dx,dy,dis2); } } printf("%I64d\n",dis[find(1)]); } return 0; }
并查集+贪心 HDU4424
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。