zoukankan      html  css  js  c++  java
  • HDU 4424 Conquer a New Region (并查集)

    题意:给定一棵树,让你选一个中心城市,问你中心城市到所有其他城市的权值和最小,每条路的权值是该边上和最小权值。

    析:从大到小枚举边,然后用并查集进行维护,在合并两个集合时,要考虑两边集合加上该边的的最大的那一个,每次要取最大值。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #include <list>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 2e5 + 10;
    const LL mod = 1e12;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    
    int p[maxn];
    int num[maxn];
    LL sum[maxn];
    struct Node{
      int u, v, w;
      bool operator < (const Node &p) const{
        return w > p.w;
      }
    };
    Node a[maxn];
    
    int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }
    
    int main(){
      while(scanf("%d", &n) == 1){
        for(int i = 0; i < n-1; ++i)  scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w);
        memset(sum, 0, sizeof sum);
        for(int i = 0; i <= n; ++i){
          p[i] = i;
          num[i] = 1;
        }
        sort(a, a + n - 1);
        for(int i = 0; i < n-1; ++i){
          int x = Find(a[i].u);
          int y = Find(a[i].v);
          LL tmp1 = sum[x] + (LL)num[y] * a[i].w;
          LL tmp2 = sum[y] + (LL)num[x] * a[i].w;
          sum[x] = max(tmp1, tmp2);
          num[x] += num[y];
          p[y] = x;
        }
        printf("%I64d
    ", sum[Find(1)]);
      }
      return 0;
    }
    

      

  • 相关阅读:
    0316复利计算
    操作系统实验0
    0302-软件工程第一次作业
    1203有穷自动机
    11.11对同学们的作业一些评论
    1029 文法分析
    编译原理第一次上机作业感想
    词法分析编译感想
    0909 初识编译原理
    0909编译原理
  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7232803.html
Copyright © 2011-2022 走看看