状态压缩dp第一题

标签： ACM

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题目：

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.

题意：

第一行输入草地长宽，后面输入该地方能不能使用，输出可以使用的所有方案

解题思路

从例题来看第一层有五种可能分别为000，001，010，100，101，都标记为1种可能

第二层可以有000与010两种状态,但是与上一层比较000与上一层五种状态都不冲突标记为5种可能，而010与上一层010状态冲突，所以标记为4种可能

第二层为最底层，将最后一层的可能性全部相加得到9

使用状态压缩，将所有可能存在状态储存到数组里面

然后从第一层存在的状态标记为1

从第二层开始遍历到最后一层，第二层存在的状态且不和上一层冲突将上一层的状态标记加到该层的标记上

遍历到最后一层时将最后一层的状态总和加起来就是所有的可能性

注：根据题意答案要对100000000取余

AC代码

#include <iostream>
#include <string.h>
#define M 4100
#define N 15
using namespace std;
int map[N]; //该行的输入状态
int m,n;
int dp[N][M];
int p;//该列最大状态
int s[M];   //储存每一行拥有的状态最大4096种状态
int mod=100000000;
bool checkLine(int i)    //该行是否满足条件
{
    return !(i&(i>>1));
}
bool checkTwoLine(int i,int j)  //与上一行是否冲突
{
    return !(i&j);
}
bool include(int i,int j)   //是否是包含关系
{
    return ((i|j)==i);
}
void init()
{
    p=0;
    int i,j;
    for(i=0;i<(1<<m);i++)
        if(checkLine(i))
            s[p++]=i;
}
void solve()
{
    int i,j,k;
    int ans=0;
    for(i=0;i<p;i++)
        if(include(map[0],s[i]))
        dp[0][i]=1;
    for(i=1;i<n;i++)
        for(j=0;j<p;j++)    //该行的状态
        {
            if(!include(map[i],s[j]))
                continue;
            else
                for(k=0;k<p;k++)    //上一行的状态
            {
                if(include(map[i-1],s[k])&&checkTwoLine(s[j],s[k]))
                    dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;
            }
        }
    for(i=0;i<p;i++)
        ans=(ans+dp[n-1][i])%mod;
    cout<<ans<<endl;
}
int main()
{
    while(cin>>n>>m)
    {
        memset(map,0,sizeof(map));
        int i,j;
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                int plant;cin>>plant;
                if(plant){
                    map[i]+=(1<<j);     //将输入转换成二进制储存
                }
            }
        init();
        solve();
    }
    return 0;
}