zoukankan      html  css  js  c++  java
  • (简单) ZOJ 3209 Treasure Map , DLX+精确覆盖。

      Description

      Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luckily, it is possible to figure out the position of each piece in the original map. Now the boss asks you, the talent programmer, to make a complete treasure map with these pieces. You need to make only one complete map and it is not necessary to use all the pieces. But remember, pieces are not allowed to overlap with each other (See sample 2).

      题目就是找到几个矩形来覆盖一个大的矩形,而且要求不能重叠。

      构造01矩阵的话就是把n*m这个大矩形的每一个点都作为一列(也就是有n*m列),然后把每一个可选择矩形都当做一行。。。

    代码如下:

    #include<iostream>
    #include<cstring>
    
    using namespace std;
    
    const int INF=10e8;
    const int MaxN=600;
    const int MaxM=33*33;
    const int MaxNode=900*500+10;
    
    struct DLX
    {
        int n,m,size;
        int U[MaxNode],D[MaxNode],L[MaxNode],R[MaxNode],col[MaxNode];
        int S[MaxM],H[MaxN];
        int ans;
    
        void init(int _n,int _m)
        {
            n=_n;
            m=_m;
    
            ans=INF;
    
            for(int i=0;i<=m;++i)
            {
                S[i]=0;
                U[i]=D[i]=i;
                L[i]=i-1;
                R[i]=i+1;
            }
            L[0]=m;
            R[m]=0;
    
            size=m;
    
            for(int i=0;i<=n;++i)
                H[i]=-1;
        }
    
        void Link(int r,int c)    //r is row , c is col.
        {
            col[++size]=c;
            ++S[c];
    
            U[size]=U[c];
            D[U[c]]=size;
            D[size]=c;
            U[c]=size;
    
            if(H[r]<0)
                H[r]=L[size]=R[size]=size;
            else
            {
                R[size]=H[r];
                L[size]=L[H[r]];
                R[L[H[r]]]=size;
                L[H[r]]=size;
            }
        }
    
        void remove(int c)
        {
            R[L[c]]=R[c];
            L[R[c]]=L[c];
    
            for(int i=D[c];i!=c;i=D[i])
                for(int j=R[i];j!=i;j=R[j])
                {
                    D[U[j]]=D[j];
                    U[D[j]]=U[j];
                    --S[col[j]];
                }
        }
    
        void resume(int c)
        {
            for(int i=U[c];i!=c;i=U[i])
                for(int j=L[i];j!=i;j=L[j])
                {
                    ++S[col[j]];
                    D[U[j]]=U[D[j]]=j;
                }
    
            R[L[c]]=L[R[c]]=c;
        }
    
        void Dance(int d)
        {
            if(R[0]==0)
            {
                if(d<ans)
                    ans=d;
    
                return;
            }
    
            if(d>ans)
                return;
    
            int c=R[0];
            
            for(int i=R[0];i!=0;i=R[i])
                if(S[i]<S[c])
                    c=i;
    
            remove(c);
    
            for(int i=D[c];i!=c;i=D[i])
            {
                for(int j=R[i];j!=i;j=R[j])
                    remove(col[j]);
    
                Dance(d+1);
    
                for(int j=L[i];j!=i;j=L[j])
                    resume(col[j]);
            }
    
            resume(c);
        }
    };
    
    DLX dlx;
    
    int main()
    {
        ios::sync_with_stdio(false);
    
        int T;
        int n,m,p;
        int x1,x2,y1,y2;
        cin>>T;
    
        while(T--)
        {
            cin>>n>>m>>p;
    
            dlx.init(p,(n)*(m));
    
            for(int i=1;i<=p;++i)
            {
                cin>>x1>>y1>>x2>>y2;
    
                for(int x=x1;x<x2;++x)
                    for(int y=y1;y<y2;++y)
                        dlx.Link(i,y*(n)+x+1);
            }
    
            dlx.Dance(0);
    
            if(dlx.ans==INF)
                cout<<-1<<endl;
            else
                cout<<dlx.ans<<endl;
        }
    
        return 0;
    }
    View Code
  • 相关阅读:
    php 1231
    php 1229
    php 1228
    php 0103
    php 1227
    php 1230
    php 0104
    flex弹性布局学习
    ps抠图的几种方法
    sql2005 不同的日期展示形式
  • 原文地址:https://www.cnblogs.com/whywhy/p/4263872.html
Copyright © 2011-2022 走看看