标签：cst bool 棋盘 blog com ons pair == algorithm
题意：给出一个n*m的棋盘，及一个小的矩形1*2，问用这个小的矩形将这个大的棋盘覆盖有多少种方法。
析：对第（i，j）位置，要么不放，要么竖着放，要么横着放，如果竖着放，我们记第 （i，j）位置为0，（i+1，j）为1，如果横着放，那么我们记
（i，j），（i，j+1）都为1，然后dp[i][s]表示 第 i 行状态为 s 时，有多少方法，那么我们就可以考虑与和匹配的状态，这个也很容易判断。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 100000000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[12][1<<11];
bool judge(int i, int j){
int cnt = 0;
for(int k = 0; k < m; ++k){
if(i&(1<<k)){
if(j&(1<<k)) cnt ^= 1;
else if(cnt) return false;
}
else{
if(!(j&(1<<k))) return false;
if(cnt) return false;
}
}
return cnt == 0;
}
bool judge1(int i){
int cnt = 0;
for(int j = 0; j < m; ++j)
if(i&(1<<j)) cnt ^= 1;
else if(cnt) return false;
return cnt == 0;
}
int main(){
while(scanf("%d %d", &n, &m) == 2 && m+n){
if(n < m) swap(n, m);
if(n * m & 1){ cout << "0" << endl; continue; }
int all = 1 << m;
memset(dp, 0, sizeof dp);
for(int i = 0; i < all; ++i)
dp[1][i] = judge1(i);
for(int i = 2; i <= n; ++i){
for(int j = 0; j < all; ++j)
for(int k = 0; k < all; ++k)
if(judge(j, k)) dp[i][j] += dp[i-1][k];
}
cout << dp[n][all-1] << endl;
}
return 0;
}
POJ 2411 Mondriaan's Dream (状压DP)
标签：cst bool 棋盘 blog com ons pair == algorithm
原文地址：http://www.cnblogs.com/dwtfukgv/p/6892193.html
