POJ 3254 - Corn Fields （状压DP）

Corn Fields Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17014   Accepted: 8987 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

题意：

给你一个地图，1可以放置。 让你求出放置的方案总数。注意放置不能上下左右相邻，放置数可以为0。

POINT：

nm才12，可以状压遍历状态来做。一层一层更新答案。

#include <string>
#include <string.h>
#include <iostream>
#include <queue>
#include <math.h>
#include <stdio.h>
using namespace std;
const int maxn = 2;
const int mod = 100000000;
int a[22][22];
int kind[1000];
int n,m;
int num=0;
void dfs()
{
for(int i=0;i<=((1<<m)-1);i++){
int x=i,flag=1;
while(x){
if(x&1&&flag==1) flag=0;
else if(x&1&&flag==0){
break;
}
else{
flag=1;
}
x>>=1;
}
if(x==0){
kind[++num]=i;
}
}
}
int dp[5000];
int main()
{
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&a[i][j]);
}
}
dfs();
for(int j=1;j<=num;j++){
int x=kind[j],flag=1;
for(int cnt=1;cnt<=m;cnt++){
if(a[1][cnt]==0&&((x>>(cnt-1))&1)==1){
flag=0;
break;
}
}
if(flag) dp[x]=1;
}
int dp1[5000];
for(int i=2;i<=n;i++){
memset(dp1,0,sizeof dp1);
for(int j=1;j<=num;j++){
for(int k=1;k<=num;k++){
int x=kind[j];
int y=kind[k];
int flag=1;
if((x&y)>0){
continue;
}
for(int cnt=1;cnt<=m;cnt++){
if(a[i][cnt]==0&&((y>>(cnt-1))&1)==1){
flag=0;
break;
}
}
if(flag){
(dp1[y]+=dp[x])%=mod;
}
}
}
for(int j=1;j<=num;j++){
dp[kind[j]]=dp1[kind[j]];
}
}
int ans=0;
for(int i=1;i<=num;i++){
(ans+=dp[kind[i]])%=mod;
}
printf("%d\n",ans);
return 0;
}