• HDU 3938 并查集


    求小于L的路径点对数(路上的最大值),按权值排序,从小到大并查集建图,有点kruskal的意思。

    /** @Date    : 2017-09-22 17:30:11
      * @FileName: HDU 3938 并查集 离线.cpp
      * @Platform: Windows
      * @Author  : Lweleth (SoungEarlf@gmail.com)
      * @Link    : https://github.com/
      * @Version : $Id$
      */
    #include <bits/stdc++.h>
    #define LL long long
    #define PII pair<int ,int>
    #define MP(x, y) make_pair((x),(y))
    #define fi first
    #define se second
    #define PB(x) push_back((x))
    #define MMG(x) memset((x), -1,sizeof(x))
    #define MMF(x) memset((x),0,sizeof(x))
    #define MMI(x) memset((x), INF, sizeof(x))
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    const int N = 1e4+20;
    const double eps = 1e-8;
    
    int n, m, q;
    struct yuu{
    	int a, b;
    	int l;
    }e[N*5];
    
    struct sion{
    	int m, l;
    }qu[N];
    
    int cmp1(yuu a, yuu b)
    {
    	return a.l < b.l;
    }
    
    int cmp2(sion a, sion b)
    {
    	return a.l < b.l;
    }
    
    int fa[N];
    LL cnt[N];
    LL ans[N];
    
    int find(int x)
    {
    	if(x != fa[x])
    		fa[x] = find(fa[x]);
    	return fa[x];
    }
    
    int join(int a, int b)
    {
    	int x = find(a);
    	int y = find(b);
    	int t = 0;
    	if(x != y)
    	{
    		t = cnt[x] * cnt[y];
    		fa[y] = x;
    		cnt[x] += cnt[y];
    		cnt[y] = 0;
    		return t;
    	}
    	return 0;
    }
    
    int main()
    {
    
    	while(~scanf("%d%d%d", &n, &m, &q))
    	{
    		MMF(ans);
    		MMF(cnt);
    
    		for(int i = 0; i <= n; i++)
    			fa[i] = i, cnt[i] = 1;
    		for(int i = 0; i < m; i++)
    			scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].l);
    		for(int i = 0; i < q; i++)
    			scanf("%d", &qu[i].l), qu[i].m = i;
    		sort(e, e + m, cmp1);
    		sort(qu, qu + q, cmp2);
    		int j = 0;
    		LL t = 0;
    		for(int i = 0; i < q; i++)
    		{
    			while(j < m && e[j].l <= qu[i].l)
    				t += join(e[j].a, e[j].b), j++;
    			ans[qu[i].m] = t;
    		}
    		//cout << "~";
    		for(int i = 0; i < q; i++)
    			printf("%lld
    ", ans[i]);
    	}
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Yumesenya/p/7583275.html
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