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题意:中文。点击打开链接个n1*m1,故将n*m作为N。那么bool值又是如何确定的了。n1*m1块的区域内的怪物对
应的第几块小区域和怪物编号就是1.接下来就是DLX做法求最小的值。
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<string> #include<iostream> #include<queue> #include<cmath> #include<map> #include<stack> #include<bitset> using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) typedef long long LL; typedef pair<int,int>pil; const int maxn=15*15+10; const int maxnnode=maxn*maxn; const int mod = 1000000007; struct DLX{ int n,m,size; int U[maxnnode],D[maxnnode],L[maxnnode],R[maxnnode],Row[maxnnode],Col[maxnnode]; int H[maxn],S[maxn];//H[i]位置,S[i]个数 int ansd; void init(int a,int b) { n=a; m=b; REPF(i,0,m) { S[i]=0; U[i]=D[i]=i; L[i]=i-1; R[i]=i+1; } R[m]=0; L[0]=m; size=m; REPF(i,1,n) H[i]=-1; } void link(int r,int c) { ++S[Col[++size]=c]; Row[size]=r; D[size]=D[c]; U[D[c]]=size; U[size]=c; D[c]=size; if(H[r]<0) H[r]=L[size]=R[size]=size; else { R[size]=R[H[r]]; L[R[H[r]]]=size; L[size]=H[r]; R[H[r]]=size; } } void remove(int c) { for(int i=D[c];i!=c;i=D[i]) L[R[i]]=L[i],R[L[i]]=R[i]; } void resume(int c) { for(int i=U[c];i!=c;i=U[i]) L[R[i]]=R[L[i]]=i; } bool v[maxn]; int f() { int ret = 0; for(int c = R[0];c != 0;c = R[c])v[c] = true; for(int c = R[0];c != 0;c = R[c]) if(v[c]) { ret++; v[c] = false; for(int i = D[c];i != c;i = D[i]) for(int j = R[i];j != i;j = R[j]) v[Col[j]] = false; } return ret; } void Dance(int d) { if(d + f() >=ansd) return ; if(R[0] == 0) { if(d<ansd) ansd=d; return ; } int c = R[0]; for(int i = R[0];i != 0;i = R[i]) if(S[i] < S[c]) c = i; for(int i = D[c];i != c;i = D[i]) { remove(i); for(int j = R[i];j != i;j = R[j])remove(j); Dance(d+1); for(int j = L[i];j != i;j = L[j])resume(j); resume(i); } } }; DLX L; int n,m; int id[25][25]; int main() { int x,n1,m1; while(~scanf("%d%d",&n,&m)) { int num=0; CLEAR(id,0); REP(i,n) { REP(j,m) { scanf("%d",&id[i][j]); if(id[i][j]) id[i][j]=(++num); } } L.init(n*m,num); num=1; scanf("%d%d",&n1,&m1); REP(i,n) { REP(j,m) { for(int xx=0;xx<n1&&i+xx<n;xx++) { for(int yy=0;yy<m1&&j+yy<=m;yy++) if(id[i+xx][j+yy]) L.link(num,id[i+xx][j+yy]); } num++; } } // cout<<"fuck "<<endl; L.ansd=0x3f3f3f3f; L.Dance(0); printf("%d\n",L.ansd); } return 0; }
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原文地址:http://blog.csdn.net/u013582254/article/details/42133755