hdu3586 Information Disturbing 【树形dp】

题目链接

题解

二分 + 简单的树形dp

我正有练一下dp的必要了

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");

using namespace std;

const int maxn = 1005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int h[maxn],ne = 2;

struct EDGE{int to,nxt,w;}ed[maxm];

inline void build(int u,int v,int w){

	ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;

	ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;

}

int fa[maxn],w[maxn];

int n,m,lim;

void DFS(int u){

	Redge(u) if ((to = ed[k].to) != fa[u]){

		fa[to] = u; w[to] = ed[k].w;

		DFS(to);

	}

}

LL f[maxn][2];

void dfs(int u){

	if (w[u] <= lim) f[u][1] = w[u];

	else f[u][1] = INF;

	f[u][0] = 0;

	int cnt = 0;

	Redge(u) if ((to = ed[k].to) != fa[u]){

		dfs(to); cnt++;

		f[u][0] += min(f[to][0],f[to][1]);

	}

	if (!cnt) f[u][0] = INF;

}

bool check(int x){

	lim = x;

	dfs(1);

	return f[1][0] <= m;

}

int main(){

	while ((n = read()) && (m = read())){

		memset(h,0,sizeof(h)); ne = 2;

		int a,b,w;

		for (int i = 1; i < n; i++){

			a = read(); b = read(); w = read();

			build(a,b,w);

		}

		DFS(1);

		int l = 1,r = m,mid;

		while (l < r){

			mid = l + r >> 1;

			if (check(mid)) r = mid;

			else l = mid + 1;

		}

		if (check(l)) printf("%d\n",l);

		else puts("-1");

	}

	return 0;

}

巴特西

hdu3586 Information Disturbing 【树形dp】

题目链接

题解

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