QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#874205#7578. Salty FishCarroT1212AC ✓900ms103176kbC++141.6kb2025-01-27 20:33:222025-01-27 20:33:23

Judging History

你现在查看的是最新测评结果

  • [2025-01-27 20:33:23]
  • 评测
  • 测评结果:AC
  • 用时:900ms
  • 内存:103176kb
  • [2025-01-27 20:33:22]
  • 提交

answer

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
using namespace std; bool MEM;
using ll=long long; using ld=long double;
using pii=pair<int,int>; using pll=pair<ll,ll>;
const int I=1e9;
const ll J=1e18,N=3e5+7;
ll n,m,fa[N],a[N],sum,ans;
ll hl[N],ls[N],dep[N];
map<ll,ll> mp[N];
vector<ll> e[N];
vector<pll> qv[N];
void dfs(ll p,ll f) {
	hl[p]=ls[p]=0,dep[p]=dep[f]+1;
	for (ll i:e[p]) if (i!=f) {
		dfs(i,p);
		if (!ls[p]||hl[i]>hl[ls[p]]) ls[p]=i,hl[p]=hl[i]+1;
	}
}
void dfs1(ll p,ll f) {
	if (ls[p]) {
		dfs1(ls[p],p);
		mp[p].swap(mp[ls[p]]);
		for (ll i:e[p]) if (i!=f&&i!=ls[p]) {
			dfs1(i,p);
			for (pll j:mp[i]) mp[p][j.fi]+=j.se;
		}
	}
	mp[p][dep[p]]+=a[p];
	for (pll o:qv[p]) {
		ll k=o.fi,w=o.se; auto it=mp[p].begin();
		while (it=mp[p].upper_bound(dep[p]+k),it!=mp[p].begin()) {
			it--;
			ll x=it->fi,y=it->se,z=min(y,w);
			y-=z,w-=z,ans+=z,mp[p][x]-=z;
			if (!y) mp[p].erase(x);
			if (!w) break;
		}
	}
//	printf("%lld\n",p);
//	for (pll i:mp[p]) printf("%lld %lld | ",i.fi,i.se); printf("\n");
}
void mian() {
	scanf("%lld%lld",&n,&m),sum=ans=0;
	for (ll i=1;i<=n;i++) e[i].clear(),qv[i].clear(),mp[i].clear();
	for (ll i=2;i<=n;i++) scanf("%lld",&fa[i]),e[fa[i]].pb(i);
	for (ll i=1;i<=n;i++) scanf("%lld",&a[i]),sum+=a[i];
	for (ll i=1,x,k,c;i<=m;i++) scanf("%lld%lld%lld",&x,&k,&c),qv[x].pb({k,c});
	dfs(1,0),dfs1(1,0);
	cout<<sum-ans<<"\n"; 
}
bool ORY; int main() {
	// while (1)
	 int t; for (scanf("%d",&t);t--;)
	mian();
	cerr<<"\n"<<abs(&MEM-&ORY)/1048576<<"MB";
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 3ms
memory: 40408kb

input:

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

output:

6

result:

ok 1 number(s): "6"

Test #2:

score: 0
Accepted
time: 900ms
memory: 103176kb

input:

1003
100 100
1 2 3 4 5 6 7 8 9 10 6 1 2 4 1 11 17 14 17 2 13 8 8 5 11 7 18 6 2 10 23 11 13 3 9 1 33 20 3 9 32 35 11 41 42 29 33 45 21 35 9 36 12 54 19 24 57 31 32 5 3 10 46 15 46 48 20 44 5 41 67 7 18 30 27 6 29 69 57 75 62 74 18 64 17 21 38 60 79 69 54 90 83 83 31 96 31 93 53
152 498 653 559 287 38...

output:

20180
17083
14650
19924
15814
20189
20894
18175
18478
13758
20217
23680
15562
12882
18046
18132
20548
17000
23734
18740
24814
16728
20979
19727
16450
21717
15739
22081
17803
23024
14820
21503
23497
15804
18097
17197
21236
21286
14250
11632
18335
12317
16313
22840
18583
15245
19331
25978
22388
17091
...

result:

ok 1003 numbers