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  • Poj1741/洛谷P4718 Tree(点分治)

    题面

    有多组数据:Poj

    无多组数据:洛谷

    题解

    点分治板子题,$calc$的时候搞一个$two pointers$扫一下统计答案就行了。

    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using std::min; using std::max;
    using std::swap; using std::sort;
    typedef long long ll;
    
    template<typename T>
    void read(T &x) {
        int flag = 1; x = 0; char ch = getchar();
        while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
        while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
    }
    
    const int N = 1e4 + 10, Inf = 1e9 + 7;
    int n, k, d[N], Size;
    int cnt, from[N], to[N << 1], nxt[N << 1], dis[N << 1];
    int p, siz[N], tmp; bool vis[N];
    inline void addEdge(int u, int v, int w) {
    	to[++cnt] = v, nxt[cnt] = from[u], dis[cnt] = w, from[u] = cnt;
    }
    
    void getrt(int u, int f) {
    	siz[u] = 1; int max_part = 0;
    	for(int i = from[u]; i; i = nxt[i]) {
    		int v = to[i]; if(v == f || vis[v]) continue;
    		getrt(v, u), siz[u] += siz[v];
    		max_part = max(max_part, siz[v]);
    	} max_part = max(max_part, Size - siz[u]);
    	if(max_part < tmp) tmp = max_part, p = u;
    }
    
    void getpoi(int x, int y, int f) {
    	d[++tmp] = y;
    	for(int i = from[x]; i; i = nxt[i]) {
    		int v = to[i]; if(v == f || vis[v]) continue;
    		getpoi(v, y + dis[i], x);
    	}
    }
    
    int calc (int x, int y) {
    	tmp = 0, getpoi(x, y, 0);
    	sort(&d[1], &d[tmp + 1]);
    	int j = tmp, ret = 0;
    	for(int i = 1; i < j; ++i) {
    		while(d[i] + d[j] > k && i < j) --j;
    		ret += j - i;
    	} return ret;
    }
    
    int doit(int x) {
    	int ret = 0; 
    	tmp = Inf, getrt(x, 0), vis[p] = true;
    	ret += calc(p, 0);
    	for(int i = from[p]; i; i = nxt[i]) {
    		int v = to[i]; if(vis[v]) continue;
    		ret -= calc(v, dis[i]), Size = siz[v], ret += doit(v);
    	} return ret;
    }
    
    int main () {
    	while(1) {
    		read(n), read(k); if(!n && !k) break;
    		Size = n, cnt = 0, memset(from, 0, sizeof from);
    		memset(vis, 0, sizeof vis);
    		for(int i = 1, u, v, w; i < n; ++i) {
    			read(u), read(v), read(w);
    			addEdge(u, v, w), addEdge(v, u, w);
    		}
    		printf("%d
    ", doit(1));
    	}
    	return 0;
    } 
    
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  • 原文地址:https://www.cnblogs.com/water-mi/p/10304060.html
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