把每一个怪物编号,然后枚举矩形左上角,扫描矩形内的怪物编号建图,行为矩形的左上角编号,列为怪物编号,模型转化为选取最少的矩形将列上的怪物覆盖,很明显的
dlx重复覆盖。
代码:
/* *********************************************** Author :_rabbit Created Time :2014/4/29 16:17:31 File Name :9.cpp ************************************************ */ #pragma comment(linker, "/STACK:102400000,102400000") #include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include <stdlib.h> #include <string.h> #include <limits.h> #include <string> #include <time.h> #include <math.h> #include <queue> #include <stack> #include <set> #include <map> using namespace std; #define INF 0x3f3f3f3f #define eps 1e-8 #define pi acos(-1.0) typedef long long ll; struct DLX{ const static int maxn=2010; #define FF(i,A,s) for(int i = A[s];i != s;i = A[i]) int L[maxn],R[maxn],U[maxn],D[maxn]; int size,col[maxn],row[maxn],s[maxn],H[maxn]; bool vis[2010]; int ans[maxn],cnt; void init(int m){ for(int i=0;i<=m;i++){ L[i]=i-1;R[i]=i+1;U[i]=D[i]=i;s[i]=0; } memset(H,-1,sizeof(H)); L[0]=m;R[m]=0;size=m+1; } void link(int r,int c){ U[size]=c;D[size]=D[c];U[D[c]]=size;D[c]=size; if(H[r]<0)H[r]=L[size]=R[size]=size; else { L[size]=H[r];R[size]=R[H[r]]; L[R[H[r]]]=size;R[H[r]]=size; } s[c]++;col[size]=c;row[size]=r;size++; } void del(int c){//精确覆盖 L[R[c]]=L[c];R[L[c]]=R[c]; FF(i,D,c)FF(j,R,i)U[D[j]]=U[j],D[U[j]]=D[j],--s[col[j]]; } void add(int c){ //精确覆盖 R[L[c]]=L[R[c]]=c; FF(i,U,c)FF(j,L,i)++s[col[U[D[j]]=D[U[j]]=j]]; } bool dfs(int k){//精确覆盖 if(!R[0]){ cnt=k;return 1; } int c=R[0];FF(i,R,0)if(s[c]>s[i])c=i; del(c); FF(i,D,c){ FF(j,R,i)del(col[j]); ans[k]=row[i];if(dfs(k+1))return true; FF(j,L,i)add(col[j]); } add(c); return 0; } void remove(int c){//重复覆盖 FF(i,D,c)L[R[i]]=L[i],R[L[i]]=R[i]; } void resume(int c){//重复覆盖 FF(i,U,c)L[R[i]]=R[L[i]]=i; } int A(){//估价函数 int res=0; memset(vis,0,sizeof(vis)); FF(i,R,0)if(!vis[i]){ res++;vis[i]=1; FF(j,D,i)FF(k,R,j)vis[col[k]]=1; } return res; } void dfs(int now,int &lim){//重复覆盖 if(R[0]==0)lim=min(lim,now); else if(now+A()<lim){ int temp=INF,c; FF(i,R,0)if(temp>s[i])temp=s[i],c=i; FF(i,D,c){ remove(i);FF(j,R,i)remove(j); dfs(now+1,lim); FF(j,L,i)resume(j);resume(i); } } } }dlx; int main() { int n,m,rr,cc,mp[20][20],x,n1,m1; while(~scanf("%d%d",&n,&m)){ cc = 0;rr = 0 ; memset(mp,0,sizeof(mp)); for(int i = 1 ; i <= n ; i ++ ) for(int j = 1 ; j <= m ; j++ ){ scanf("%d",&x); if(x == 1 )mp[i][j] = ++ cc; else mp[i][j] = 0 ; } dlx.init(cc); scanf("%d%d",&n1,&m1); for(int i = 1 ; i<= n - n1 + 1 ; i ++ ) for(int j = 1 ; j <= m - m1 + 1; j ++ ) { rr ++ ; for(int i1 = i ; i1 <= i + n1 - 1 ; i1 ++ ) for(int j1 = j ; j1 <= j + m1 - 1 ; j1 ++ ) if(mp[i1][j1] != 0 ) dlx.link(rr,mp[i1][j1]); } int ans = n * m; dlx.dfs(0,ans); printf("%d\n",ans); } return 0; }
FZU 1686 DLX建图重复覆盖,码迷,mamicode.com
原文地址:http://blog.csdn.net/xianxingwuguan1/article/details/24724375