POJ3254——Corn Fields（状态压缩DP）

最新推荐文章于 2021-08-11 14:43:11 发布

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最新推荐文章于 2021-08-11 14:43:11 发布

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分类专栏：动态规划

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本文介绍了USACO 2006 November Gold比赛中的Corn Fields问题，探讨了如何运用状态压缩动态规划（DP）来解决这个问题。文章提到作者在准备省选期间遇到的挑战，包括曾经因为一个未发现的bug导致的问题，最终成功修复并解决了这个问题。

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Corn Fields

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 4830 Accepted: 2551

Description

Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.
Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.

Input
Line 1: Two space-separated integers: M and N
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)
Output
Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.
Sample Input
2 3
1 1 1
0 1 0
Sample Output
9
Hint
Number the squares as follows:
1 2 3
4

There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.
Source

USACO 2006 November Gold

解析：

省选即将到来。。。又回来搞状压DP

这是很久前写的，当时有个bug一直没找到错，特判哈也过了。。。

今天把bug找出来重新A。。。

很简单的状压。。。转移时只用判断每行是否可行和相邻行是否可行。。。

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

int n,m,tol;
bool vis[1<<13][1<<13];
long long f[13][1<<13];
int map[13][13];
long long ans=0;

void readdata()
{
    freopen("poj3254.in","r",stdin);
    freopen("poj3254.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        scanf("%d",&map[i][j]);
    tol=(1<<m)-1;
}

bool check(int s1,int s2)  //s2-->s1是否可行
{
    int i,x,j,k,l;
    for(i=0;i<m;i++)
    {
        x=(s1>>i)&1;
        j=(s2>>i)&1;
        if((x&j)==1)return 0;
    }
    return 1;
}

bool ok(int x)//判断当前状态是否可行（不能相邻）
{
    if(x&(x<<1))return 0;
    return 1;
}

bool can(int i,int s)
{
    for(int j=1;j<=m;j++)
    {
        if(map[i][j]==0 && (s>>(m-j))&1)return 0;
    }
    return 1;
}

void init()
{
    memset(f,0,sizeof(f));
    for(int i=0;i<=tol;i++)
    {
        if(can(1,i))f[1][i]=1;
    }
    for(int i=0;i<=tol;i++)
        for(int j=0;j<=tol;j++)
            if(ok(i) && ok(j) && check(i,j))vis[i][j]=true;//相邻状态可行就标记为true
}

void work()
{
    for(int i=2;i<=n;i++)
        for(int j=0;j<=tol;j++)
            for(int k=0;k<=tol;k++)
            {
                if(vis[j][k] && can(i,j) && can(i-1,k))
                    f[i][j]+=f[i-1][k];
            }
    for(int i=0;i<=tol;i++)
    {
        if(ok(i))ans+=f[n][i];
        ans%=100000000;
    }
}

int main()
{
    readdata();
    init();
    work();
    cout<<ans<<endl;
    return 0;
}