题意:又是farmer john和他的牛,给定一个M*N的玉米地,有的点上有玉米,有的则没有,问把这些牛放在有玉米的点上且不相邻总共有多少种方法,牛的个数可以放0个,1个以及一个图上不相邻点的最大团都可以。
题解:动态规划。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxP 100000000
class solve
{
private:
int M,N;
int fertile[13];
int num[13][378];
int feasibleState[378];
int stateNum;
public:
solve(int m,int n):M(m),N(n)
{
processIn();
generateState();
printf("%d\n",dp());
}
int processIn();
bool IsFeasible(int state);
int generateState();
int dp();
};
int solve::dp()
{
int i,j,k;
int oldState,newState;
memset(num,0,sizeof(num));
num[0][0] = 1;
for(i = 1;i <= M;i++)
{
for(j = 0;j < stateNum;j++)
{
oldState = feasibleState[j];
for(k = 0;k < stateNum;k++)
{
newState = feasibleState[k];
if((newState&fertile[i]) == newState&&(oldState&newState) == 0)
{
num[i][k] = (num[i-1][j]+num[i][k])%maxP;
}
}
}
}
int totNum = 0;
for(i = 0;i < stateNum;i++)
{
totNum = (totNum+num[M][i])%maxP;
}
return totNum;
}
bool solve::IsFeasible(int state)
{
while(state)
{
if((state&3) == 3)
return 0;
state >>= 1;
}
return 1;
}
int solve::generateState()
{
int i,j;
stateNum = 0;
j = 1<<N;
for(i = 0;i < j;i++)
{
if(IsFeasible(i))
{
feasibleState[stateNum++] = i;
}
}
return 0;
}
int solve::processIn()
{
int i,j;
int f;
memset(fertile,0,sizeof(fertile));
for(i = 1;i <= M;i++)
{
for(j = 0;j < N;j++)
{
scanf("%d",&f);
fertile[i] <<= 1;
fertile[i] |= f;
}
}
return 0;
}
int main()
{
int m,n;
while(~scanf("%d%d",&m,&n))
{
solve poj_3524(m,n);
}
return 0;
}
因篇幅问题不能全部显示,请点此查看更多更全内容