bzoj2809: [Apio2012]dispatching

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2809

题意：中文题。。

分析：稍微分析下其实就知道要从叶子往根贪心保留薪水最少的人，因为只要算人头数和根的领导能力，所以从叶子往根合并是发现花费超过了总预算就可以删除薪水高的忍者。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=100010;
const int MAX=151;
const int MOD=1000000007;
const int MOD1=100000007;
const int MOD2=100000009;
const int INF=2100000000;
const double EPS=0.00000001;
typedef long long ll;
typedef unsigned long long uI64;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll ans=0;
int m,c[N],l[N];
int tot,d[N],h[N],pre[N];
int lc[N],rc[N],fa[N],sum[N],peo[N];
void add(int a,int b) {
    d[tot]=b;pre[tot]=h[a];h[a]=tot++;
}
int find_fa(int x) {
    if (fa[x]==x) return x;
    return fa[x]=find_fa(fa[x]);
}
int mergee(int a,int b) {
    if (!a||!b) return a+b;
    if (c[a]<c[b]) { a^=b;b^=a;a^=b; }
    rc[a]=mergee(rc[a],b);
    if (h[lc[a]]<h[rc[a]]) { lc[a]^=rc[a];rc[a]^=lc[a];lc[a]^=rc[a]; }
    h[a]=h[rc[a]]+1;
    return a;
}
void dfs(int x) {
    int i,a,b,f;
    for (i=h[x];i!=-1;i=pre[i]) {
        dfs(d[i]);
        a=find_fa(x);
        b=find_fa(d[i]);
        f=mergee(a,b);
        fa[a]=fa[b]=fa[f]=f;
        sum[f]=sum[a]+sum[b];peo[f]=peo[a]+peo[b];
        while (sum[f]>m) {
            fa[lc[f]]=fa[rc[f]]=mergee(lc[f],rc[f]);
            fa[f]=fa[lc[f]];
            sum[fa[f]]=sum[f]-c[f];peo[fa[f]]=peo[f]-1;
            f=fa[f];
        }
    }
    f=find_fa(x);
    ans=max(ans,(ll)peo[f]*l[x]);
}
int main()
{
    int i,n,x;
    scanf("%d%d", &n, &m);
    tot=0;
    memset(h,-1,sizeof(h));
    for (i=1;i<=n;i++) {
        scanf("%d%d%d", &x, &c[i], &l[i]);
        add(x,i);
    }
    lc[0]=rc[0]=fa[0]=sum[0]=peo[0]=0;
    for (i=1;i<=n;i++) {
        lc[i]=rc[i]=0;fa[i]=i;sum[i]=c[i];peo[i]=1;
    }
    dfs(1);
    printf("%lld\n", ans);
    return 0;
}


/*
5 4
0 3 3
1 3 5
2 2 2
1 2 4
2 3 1

7 6
0 3 4
1 3 7
2 2 5
2 2 5
1 5 4
5 3 1
5 2 5

*/