一个数有多少位1
for(int i = 1;i <= 1024;i ++) w[i] = w[i ^ i & -i] + 1;
当状态选择有条件时,不一定要考虑从开始(111111111)局面选到当前局面的最优解,可以考虑从当前局面开始选择时的最优解。例如 https://ac.nowcoder.com/acm/contest/9985/E
dp【x】为当前剩余石子为x时先手比后说多的最多解
#include <bits/stdc++.h>
using namespace std;
const int Mn = 2e6 + 5;
#define ll long long
ll dp[Mn];
ll a[100];
int e[100];
const ll inf = 0x8080808080808080;
int main()
{
int _;scanf("%d",&_);
while(_--){
int n;scanf("%d",&n);
for(int i = 0;i < n;i ++) scanf("%lld",a + i);
int x,y;
memset(e,0,sizeof(e));
for(int i = 1;i < n;i ++){
scanf("%d%d",&x,&y);
x --;
y --;
e[x] ^= (1 << y);
e[y] ^= (1 << x);
}
memset(dp,128,sizeof(dp));
for(int i = 0;i < n;i ++){
dp[(1 << i)] = a[i];
}
dp[0] = 0;
for(int i = 0;i <= (1 << n) - 1;i ++){
for(int j = 0;j < n;j ++){
if(dp[i] == inf) continue;
if(!(i & (1 << j)) && (e[j] & i)){
dp[i^(1<<j)] = max(dp[i ^ (1<<j)],a[j] - dp[i]);
}
}
}
printf("%lld\n",dp[(1 << n) - 1]);
}
return 0;
}