[HDU 4739]Zhuge Liang's Mines:状压DP
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题目中只有20个点,因此可以枚举每种选点方案,然后进行状态压缩DP。f[i]=max(f[i],f[i^t]+4)
(我因为判断函数写错,WA了好几次)
/*User:SmallLanguage:C++Problem No.:4739*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;int n,ans,dp[(1<<20)+5];vector<int> g[25];struct no{ int x,y; bool operator<(const no b)const{ return x==b.x?y<b.y:x<b.x; }}p[25];bool ok(int i,int j,int k,int l){ if(p[i].x!=p[j].x||p[i].y!=p[k].y||p[j].y!=p[l].y||p[k].x!=p[l].x) return 0; if(abs(p[j].y-p[i].y)!=abs(p[k].x-p[i].x)) return 0; return 1;}void solve(){ ans=0; for(int i=1;i<=n;i++) g[i].clear(); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) cin>>p[i].x>>p[i].y; sort(p+1,p+n+1); for(int a=1;a<=n;a++){ for(int b=a+1;b<=n;b++){ for(int c=b+1;c<=n;c++){ for(int d=c+1;d<=n;d++){ if(ok(a,b,c,d)){ int tmp=0; tmp|=1<<(a-1); tmp|=1<<(b-1); tmp|=1<<(c-1); tmp|=1<<(d-1); g[a].push_back(tmp); } } } } } for(int i=0;i<(1<<n);i++){ for(int j=1;j<=n;j++){ if((i>>(j-1))&1){ for(int k=0;k<g[j].size();k++){ int t=g[j][k]; if((i|t)==i) dp[i]=max(dp[i^t]+4,dp[i]); } } } } cout<<dp[(1<<n)-1]<<endl;}int main(){ freopen("data.in","r",stdin);// ios::sync_with_stdio(false); while(cin>>n&&n!=-1) solve(); return 0;}
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