Hdu3498-whosyourdaddy(精确覆盖模板题)

最新推荐文章于 2019-02-20 09:23:27 发布

转载最新推荐文章于 2019-02-20 09:23:27 发布 · 76 阅读

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原文链接：http://www.cnblogs.com/wust-ouyangli/p/5747090.html

本文介绍了一个基于Warcraft游戏背景的精确覆盖问题求解方法，通过使用DLX算法优化来找出最少攻击次数消灭所有敌方单位的方案。

Problem Description

sevenzero liked Warcraft very much, but he haven't practiced it for several years after being addicted to algorithms. Now, though he is playing with computer, he nearly losed and only his hero Pit Lord left. sevenzero is angry, he decided to cheat to turn defeat into victory. So he entered "whosyourdaddy", that let Pit Lord kill any hostile unit he damages immediately. As all Warcrafters know, Pit Lord masters a skill called Cleaving Attack and he can damage neighbour units of the unit he attacks. Pit Lord can choice a position to attack to avoid killing partial neighbour units sevenzero don't want to kill. Because sevenzero wants to win as soon as possible, he needs to know the minimum attack times to eliminate all the enemys.

Input

There are several cases. For each case, first line contains two integer N (2 ≤ N ≤ 55) and M (0 ≤ M ≤ N*N),and N is the number of hostile units. Hostile units are numbered from 1 to N. For the subsequent M lines, each line contains two integers A and B, that means A and B are neighbor. Each unit has no more than 4 neighbor units. The input is terminated by EOF.

Output

One line shows the minimum attack times for each case.

Sample Input

5 4

1 2

1 3

2 4

4 5

6 4

1 2

1 3

1 4

4 5

Sample Output

2

3

解析：裸的精确覆盖问题，不过要加优化，不过也是模板。

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int INF=1e9+7;
const int ms=60;
const int maxn=ms*ms;
int N,M,ans;
struct DLX
{
    int n,id;
    int L[maxn],R[maxn],U[maxn],D[maxn];
    int C[maxn],S[maxn],loc[maxn][2];
    void init(int nn=0) //传列长
    {
        n=nn;
        for(int i=0;i<=n;i++) U[i]=D[i]=i,L[i]=i-1,R[i]=i+1;
        L[0]=n; R[n]=0;
        id=n;
        memset(S,0,sizeof(S));
    }
    void AddRow(int x,int col,int A[]) //传入参数是行标号，列长，列数组
    {
        bool has=false;
        int first=id+1;
        for(int y=1;y<=col;y++)
        {
            if(A[y]==0) continue;
            has=true;
            ++id;
            L[id]=id-1; R[id]=id+1;
            D[id]=y; U[id]=U[y];
            D[U[y]]=id; U[y]=id;
            loc[id][0]=x,loc[id][1]=y;
            C[id]=y; S[y]++;
        }
        if(!has) return;
        R[id]=first; L[first]=id;
    }
    void Remove(int Size)
    {
        for(int j=D[Size];j!=Size;j=D[j])//将左右两边连接
            L[R[j]]=L[j],R[L[j]]=R[j];
    }
    void Resume(int Size)
    {
        for(int j=U[Size];j!=Size;j=U[j])//恢复
            L[R[j]]=R[L[j]]=j;
    }
    bool vis[ms];//标记行是否访问过
    int h() //启发式函数
    {
        int ret=0;
        int i,j,k;
        memset(vis,0,sizeof(vis));
        for(i=R[0];i;i=R[i])
        {
           if(vis[i]) continue;
           ret++;
           for(j=D[i];j!=i;j=D[j]) //所有关联的标记了
               for(k=R[j];k!=j;k=R[k]) vis[C[k]]=1;
        }
        return ret;
    }
    void dfs(int step)
    {
        if(step+h()>=ans) return;
        if(R[0]==0){ ans=min(ans,step); return; }
        int Min=INF,c=-1;
        for(int i=R[0];i;i=R[i]) if(Min>S[i]){ Min=S[i]; 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);
            dfs(step+1);
            for(int j=L[i];j!=i;j=L[j]) Resume(j);
            Resume(i);
        }
        return;
    }
}dlx;
int Mart[ms][ms];
int main()
{
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        dlx.init(N);
        ans=55;
        memset(Mart,0,sizeof(Mart));
        for(int i=1;i<=N;i++) Mart[i][i]=1;
        while(M--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            Mart[x][y]=Mart[y][x]=1;
        }
        for(int i=1;i<=N;i++) dlx.AddRow(i,N,Mart[i]);
        dlx.dfs(0);
        printf("%d\n",ans);
    }
    return 0;
}

View Code

代码

转载于:https://www.cnblogs.com/wust-ouyangli/p/5747090.html