標籤： ACM

題目：

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.

題意：
第一行輸入草地長寬，後面輸入該地方能不能使用，輸出可以使用的所有方案

解題思路
從例題來看第一層有五種可能分別為000，001，010，100，101，都標記為1種可能
第二層可以有000與010兩種狀態,但是與上一層比較000與上一層五種狀態都不衝突標記為5種可能，而010與上一層010狀態衝突，所以標記為4種可能
第二層為最底層，將最後一層的可能性全部相加得到9

使用狀態壓縮，將所有可能存在狀態儲存到數組裡面
然後從第一層存在的狀態標記為1
從第二層開始遍歷到最後一層，第二層存在的狀態且不和上一層衝突將上一層的狀態標記加到該層的標記上
遍歷到最後一層時將最後一層的狀態總和加起來就是所有的可能性
注：根據題意答案要對100000000取餘

AC程式碼

#include <iostream>
#include <string.h>
#define M 4100
#define N 15
using namespace std;
int map[N]; //該行的輸入狀態
int m,n;
int dp[N][M];
int p;//該列最大狀態
int s[M];   //儲存每一行擁有的狀態最大4096種狀態
int mod=100000000;
bool checkLine(int i)    //該行是否滿足條件
{
    return !(i&(i>>1));
}
bool checkTwoLine(int i,int j)  //與上一行是否衝突
{
    return !(i&j);
}
bool include(int i,int j)   //是否是包含關係
{
    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++)    //該行的狀態
        {
            if(!include(map[i],s[j]))
                continue;
            else
                for(k=0;k<p;k++)    //上一行的狀態
            {
                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);     //將輸入轉換成二進位制儲存
                }
            }
        init();
        solve();
    }

    return 0;
}