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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 |
---|---|---|---|---|---|---|---|---|---|
#381930 | #7588. Monster Hunter | chenxinyang2006 | RE | 1ms | 7960kb | C++14 | 2.3kb | 2024-04-07 22:12:56 | 2024-04-07 22:12:57 |
Judging History
answer
#include <bits/stdc++.h>
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
//#define mod 998244353
//#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;
template <class T>
void chkmax(T &x,T y){
if(x < y) x = y;
}
template <class T>
void chkmin(T &x,T y){
if(x > y) x = y;
}
inline int popcnt(int x){
return __builtin_popcount(x);
}
inline int ctz(int x){
return __builtin_ctz(x);
}
/*ll power(ll p,int k = mod - 2){
ll ans = 1;
while(k){
if(k % 2 == 1) ans = ans * p % mod;
p = p * p % mod;
k /= 2;
}
return ans;
}*/
int T,n;
ll a[100005],b[100005];
int anc[100005];
int cnt;
int head[100005];
struct eg{
int to,nxt;
}edge[200005];
void make(int u,int v){
edge[++cnt].to = v;
edge[cnt].nxt = head[u];
head[u] = cnt;
}
void dfs(int u,int f){
anc[u] = f;
for(int i = head[u];i;i = edge[i].nxt){
int v = edge[i].to;
if(v == f) continue;
dfs(v,u);
}
}
int fa[100005];
int fnd(int u){
if(fa[u] == u) return u;
return fa[u] = fnd(fa[u]);
}
set <pll> P,Q;
void solve(){
scanf("%d",&n);
rep(u,2,n) scanf("%lld%lld",&a[u],&b[u]);
cnt = 0;
fill(head,head + n + 1,0);
rep(i,1,n - 1){
int u,v;
scanf("%d%d",&u,&v);
make(u,v);make(v,u);
}
dfs(1,0);
P.clear();Q.clear();
rep(u,1,n){
fa[u] = u;
if(u == 1) continue;
if(a[u] <= b[u]) P.insert(mkp(a[u],u));
else Q.insert(mkp(b[u],u));
}
rep(pp,1,n - 1){
int u = -1;
if(!P.empty()){
auto it = P.begin();
u = it->second;
P.erase(it);
}else{
auto it = prev(Q.end());
u = it->second;
Q.erase(it);
}
int v = fnd(anc[u]);
if(v != 1){
if(a[v] <= b[v]) P.erase(mkp(a[v],v));
else Q.erase(mkp(b[v],v));
}
ll ta = max(a[v],a[v] - b[v] + a[u]),tb = b[u] + b[v] - a[u] - a[v] + ta;
a[v] = ta;b[v] = tb;
if(v != 1){
if(a[v] <= b[v]) P.erase(mkp(a[v],v));
else Q.erase(mkp(b[v],v));
}
fa[u] = v;
}
printf("%lld\n",a[1]);
}
int main(){
scanf("%d",&T);
while(T--) solve();
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 7960kb
input:
1 4 2 6 5 4 6 2 1 2 2 3 3 4
output:
3
result:
ok 1 number(s): "3"
Test #2:
score: -100
Runtime Error
input:
10 50000 112474 198604 847262 632597 871374 962220 971398 284540 10360 695245 43233 569941 64019 203468 598469 997911 41841 657552 333198 936119 546370 180011 58831 901040 959767 396595 710441 277461 429299 209831 494164 138327 393982 581911 993164 617488 108113 160310 55841 611360 301395 291074 149...