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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 |
---|---|---|---|---|---|---|---|---|---|
#620576 | #2443. Dense Subgraph | Lavender_Field | AC ✓ | 281ms | 14024kb | C++20 | 4.2kb | 2024-10-07 19:22:27 | 2024-10-07 19:22:28 |
Judging History
answer
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i = a; i <= b; ++i)
#define pb push_back
using namespace std;
inline int rd() {
int r = 0; bool w = false; char ch = getchar();
while( ch < '0' || ch > '9' ) w = !(ch^45), ch = getchar();
while( ch >= '0' && ch <= '9' ) r = (r<<1) + (r<<3) + (ch^48), ch = getchar();
return w ? -r : r;
}
#define MAXN 35000
const int mod = 1e9+7;
const int qpow3[] = {1, 3, 9, 27, 81, 243, 729};
int n, x;
int a[MAXN+5];
bool minimal_status[MAXN+5][32];
vector<int> to[MAXN+5], son[MAXN+5]; int fa[MAXN+5];
vector<int> constraint[MAXN+5];
void dft( int u ) {
for(auto v: to[u]) if( fa[u] != v ) {
fa[v] = u;
son[u].pb(v);
dft(v);
}
}
inline void add( int u , int v ) {
to[u].pb(v);
}
int f[MAXN+5], g[MAXN+5], h[MAXN+5];
bool chk( int u , int typ , int sta ) {
if( typ == 0 ) return true;
int tmpsta = sta;
for(int j = 0; j < son[u].size(); ++j) {
if( tmpsta % 3 == 2 ) return false;
tmpsta /= 3;
}
if( typ == 1 ) {
for(auto st: constraint[u]) {
bool flag = false;
for(int j = 0; j < son[u].size(); ++j) {
if( (st>>j & 1) == 1 && sta / qpow3[j] % 3 == 0 ) {
flag = true;
break;
}
}
if( flag == false ) return false;
}
} else {
bool typ2_topflag = false;
for(auto st: constraint[u]) {
if( (st >> son[u].size()) & 1 ) {
bool flag = false;
for(int j = 0; j < son[u].size(); ++j) {
if( (st >> j & 1) == 1 && sta / qpow3[j] % 3 == 0 ) {
flag = true;
break;
}
}
if( flag == false ) typ2_topflag = true;
} else {
bool flag = false;
for(int j = 0; j < son[u].size(); ++j) {
if( (st >> j & 1) == 1 && sta / qpow3[j] % 3 == 0 ) {
flag = true;
break;
}
}
if( flag == false ) return false;
}
}
if( !typ2_topflag ) return false;
}
return true;
}
void dfs( int u ) {
if( son[u].size() == 0 ) {
f[u] = g[u] = 1;
return;
}
for(auto v: son[u]) dfs(v);
for(int s = 0; s < qpow3[son[u].size()]; ++s) {
int mul = 1;
int tmps = s;
for(int j = 0; j < son[u].size(); ++j) {
int v = son[u][j];
mul = 1ll * mul * (tmps%3 == 0 ? f[v] : (tmps%3 == 1 ? g[v] : h[v])) % mod;
tmps /= 3;
}
if( chk(u,0,s) ) f[u] = (f[u] + mul) % mod;
if( chk(u,1,s) ) g[u] = (g[u] + mul) % mod;
if( chk(u,2,s) ) h[u] = (h[u] + mul) % mod;
}
// printf("%d: %d %d %d\n", u, f[u], g[u], h[u]);
}
int popcount( int x ) {
int ret = 0;
while( x ) ret += (x&1), x >>= 1;
return ret;
}
void solve() {
n = rd(), x = rd();
FOR(i,1,n) a[i] = rd() - x;
FOR(i,1,n-1) {
int u = rd(), v = rd();
add(u,v), add(v,u);
}
dft(1);
FOR(i,1,n) {
for(int s = 1; s < 1 << to[i].size(); ++s) {
if( minimal_status[i][s] ) continue;
int sum = a[i];
if( i != 1 && ((s>>(to[i].size()-1)) & 1) ) sum += a[fa[i]];
for(int j = 0; j < son[i].size(); ++j) {
if( s>>j & 1 ) sum += a[son[i][j]];
}
if( sum > 0 ) {
for(int S = s; S < 1 << to[i].size(); S = (S + 1) | s) {
minimal_status[i][S] = true;
}
if( i == 1 || ((s >> son[i].size()) & 1) == 0 || popcount(s) != 1 )
constraint[i].pb(s);
}
}
}
dfs(1);
printf("%d\n", (f[1] + g[1]) % mod);
FOR(i,1,n) to[i].clear(), son[i].clear(), fa[i] = 0, constraint[i].clear(), memset(minimal_status[i], 0, sizeof(minimal_status[0]));
FOR(i,1,n) f[i] = g[i] = h[i] = 0;
return;
}
int main() {
int T = rd(); while( T-- ) solve();
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 281ms
memory: 14024kb
input:
30 10 11086 10189 24947 2265 9138 27104 12453 15173 3048 30054 2382 8 1 1 4 5 10 10 4 3 5 2 10 9 7 6 10 7 1 15 9664 4127 24649 13571 8586 34629 8644 3157 33133 3713 32646 29412 8108 13583 21362 23735 14 9 7 1 15 12 10 15 2 6 3 11 9 1 1 11 6 12 4 10 13 15 8 15 12 11 5 3 20 29310 21738 9421 8412 4617 ...
output:
320 3312 1048576 60461799 663660496 831386303 528321417 945912389 820338255 58962594 643816787 354510769 532631871 280169661 533022884 475656636 892230988 381315031 40006857 376652471 288541404 546717513 976236489 615823216 419541488 840899550 546169620 634600456 825179565 169850560
result:
ok 30 lines