做法:状压dp
思路
每个点与附近连边的点用二进制状态表示,1表示有边,0表示无边
该状态选取的点也用二进制状态表示,1表示选了,0表示没选
选取该点的值后的差值为该点的值减去转移前的差值
代码
// Problem: 树上博弈 // Contest: NowCoder // URL: https://ac.nowcoder.com/acm/contest/9985/E // Memory Limit: 524288 MB // Time Limit: 6000 ms // // Powered by CP Editor (https://cpeditor.org) #include <bits/stdc++.h> using namespace std; #define pb push_back #define mp(aa,bb) make_pair(aa,bb) #define _for(i,b) for(int i=(0);i<(b);i++) #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define per(i,b,a) for(int i=(b);i>=(a);i--) #define mst(abc,bca) memset(abc,bca,sizeof abc) #define X first #define Y second #define lowbit(a) (a&(-a)) #define debug(a) cout<<#a<<":"<<a<<"\n" typedef long long ll; typedef pair<int,int> pii; typedef unsigned long long ull; typedef long double ld; const int N=21; const ll INF=1e18; const int mod=1e9+7; const double eps=1e-6; const double PI=acos(-1.0); ll a[N],e[N],dp[1<<N]; void solve(){ int n;cin>>n; rep(i,0,n-1) cin>>a[i],e[i]=0; _for(i,(1<<n)) dp[i]=-INF; rep(i,1,n-1){ int u,v;cin>>u>>v; u--,v--; e[u]|=(1<<v); e[v]|=(1<<u); } rep(i,0,n-1) dp[1<<i]=a[i]; rep(i,1,(1<<n)){ if(dp[i]!=-INF){ rep(j,0,n-1){ if(!(i&(1<<j))&&(i&e[j])) dp[i|(1<<j)]=max(dp[i|(1<<j)],a[j]-dp[i]); } } } cout<<dp[(1<<n)-1]<<"\n"; } int main(){ ios::sync_with_stdio(0);cin.tie(0); int t;cin>>t;while(t--) solve(); return 0; }