BZOJ2087 : [Poi2010]Sheep

一条边能连上当且仅当它没有经过任何点，并且两边的点都是偶数个。

枚举原点，通过极角排序求出哪些边是合法的，然后区间DP即可。

时间复杂度$O(nm\log m+n^3)$。

#include<cstdio>
#include<algorithm>
const int N=610,M=20010;
int n,m,P,cnt,i,j,k,t,now,f[N][N];bool g[N][N],v[N][N];
struct vec{int x,y,p;vec(){}vec(int _x,int _y){x=_x,y=_y;}vec operator-(const vec&b){return vec(x-b.x,y-b.y);}
}a[N],b[M],c[N+M],O;
inline int cross(vec a,vec b){return a.x*b.y-a.y*b.x;}
inline bool cmp(vec x,vec y){return cross(x-O,y-O)<0;}
int dp(int l,int r){if(l+1==r)return 1;if(v[l][r])return f[l][r];v[l][r]=1;int&t=f[l][r];for(int i=l+1;i<r;i++)if(g[i][l]&&g[i][r])t=(t+dp(l,i)*dp(i,r))%P;return t;
}
int main(){scanf("%d%d%d",&n,&m,&P);for(i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);for(i=1;i<=m;i++)scanf("%d%d",&b[i].x,&b[i].y);O=a[1];std::sort(a+2,a+n+1,cmp);for(i=1;i<=n;i++)a[i].p=i;for(i=1;i<=n;i++){O=a[i];cnt=0;for(j=1;j<=n;j++)if(j!=i)c[++cnt]=a[j];for(j=1;j<=m;j++)c[++cnt]=b[j];std::sort(c+1,c+cnt+1,cmp);for(j=1,t=0;j<=cnt;j=k){for(now=0,k=j;k<=cnt&&!cross(c[k]-O,c[j]-O);k++)if(!c[k].p)now++;if(!now&&!t)for(k=j;k<=cnt&&!cross(c[k]-O,c[j]-O);k++)if(c[k].p)g[i][c[k].p]=g[c[k].p][i]=1;t^=now&1;}}return printf("%d",dp(1,n)),0;
}

转载于:https://www.cnblogs.com/clrs97/p/5939811.html

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