hdu3586 Information Disturbing 树形dp

Posted 2020-11-03 Mychael

题目链接

hdu3586

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;
}