HDU 6634 网络流最小割模型 启发式合并

如果我们先手拿完所有苹果再去考虑花费的话。

S -> 摄像头 -> 苹果 -> T

就相当于找到一个最小割使得S和T分开。

ans = sum - flow。

然后对于这一个模型， 我们可以不用网络流去解决。

我们从叶子出发，然后从下往上合并。

每次到一个节点的时候，我们先把摄像机所对应的影响去除。

然后把这个点的剩下流量传给父亲。

 

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 3e5 +100;
//vector<int> vc[N];
vector<pll> camera[N];
map<int, LL> mp[N];
map<int, LL>::iterator it;
int deep[N];
int a[N], f[N];
LL sum;
void Merge(map<int, LL> & m1, map<int, LL> & m2){
    if(m1.size() < m2.size()){
        m1.swap(m2);
    }
    for(auto & it : m2){
        m1[it.fi] += it.se;
    }
}
map<int,int> mp1, mp2;
int main(){
    int T, n, m;
    scanf("%d", &T);
    deep[1] = 1;
    while(T--){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i) mp[i].clear(), camera[i].clear();
        for(int i = 2; i <= n; ++i){
            scanf("%d", &f[i]);
            deep[i] = deep[f[i]] + 1;
        }
        sum = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d", &a[i]);
            sum += a[i];
        }
        for(int i = 1, x, k, c; i <= m; ++i){
            scanf("%d%d%d", &x, &k, &c);
            camera[x].pb({k, c});
        }
        for(int i = n; i >= 1; --i){
            mp[i][-deep[i]] += a[i];
            for(auto & t : camera[i]){
                it = mp[i].lower_bound(-(deep[i]+t.fi));
                while(it != mp[i].end()){
                    if(t.se < (it->se)){
                        sum -= t.se;
                        it -> se -= t.se;
                        t.se = 0;
                        break;
                    }
                    else {
                        sum -= it->se;
                        t.se -= it->se;
                        it = mp[i].erase(it);
                    }
                }
            }
            if(f[i]) Merge(mp[f[i]], mp[i]);
        }
        printf("%lld\n", sum);
    }
    return 0;
}