POJ 1185

 

这道题很犀利，从这道题真正了解到状态压缩DP和位运算。

同时这道题也用到了滚动数组。

着篇解题报告很好：http://apps.hi.baidu.com/share/detail/14572384

 

代码：

 

 

#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <cstring>using namespace std;void MyScanf(int &num){char a; while(a=getchar(),!isdigit(a)); num = a-'0'; while ((a = getchar())!=' ' && a!='/n') num = num*10 + a-'0';}int a[101];int n,m;int status[101];int num[101];int status_num;int f[2][61][61];int ToInt(char buf[]){int i;int result = 0;for (i = 0;i < m;i++)result = result * 2 + (buf[i] == 'H');return result;}void Input(){char buf[1001];int i;for (i = 1;i <= n;i++){scanf("%s",buf);a[i] = ToInt(buf);}}void GetAllStatus(){int i;int temp;status_num = 0;for (i = 0;i < (1 << m);i++){temp = i;if (temp & (temp << 1) || temp & (temp << 2)) continue;status[++status_num] = i;num[status_num] = 0;while (temp > 0){num[status_num] += temp % 2;temp /= 2;}}}void Init(){GetAllStatus();memset(f,0,sizeof(f));}void Print(int x){int i;for (i = 1;i <= m;i++){cout << x % 2 << " ";x /= 2;}}void Compute(){int i,j,k,p,now;now = 0;for (i = 1;i <= status_num;i++)if (status[i] & a[1])f[now][i][0] = 0;elsef[now][i][0] = num[i];//Print(a[1]);//cout << endl;////for (i = 1;i <= status_num;i++)//{//cout << status[i] << " : ";//Print(status[i]);//cout << ": " << f[now][i][0] << endl;//}for (p = 2;p <= n;p++){now = 1 - now;for (i = 1;i <= status_num;i++)for (j = 1;j <= status_num;j++){if (status[i] & a[p] || status[i] & status[j])f[now][i][j] = 0;else{if (p == 2){f[now][i][j] = num[i] + f[1 - now][j][0];//Print(status[i]);//cout << endl;//Print(status[j]);//cout << endl;//cout << f[now][i][j] << endl << endl;}else{for (k = 1;k <= status_num;k++)if (status[i] & status[k])continue;elsef[now][i][j] = max(f[now][i][j],f[1 - now][j][k] + num[i]);}}}}int result = 0;for (i = 1;i <= status_num;i++)for (j = 1;j <= status_num;j++){if (n == 1)result = max(result,f[now][i][0]);elseresult = max(result,f[now][i][j]);}cout << result << endl;}int main(){while (cin >> n >> m){Input();Init();//cout << status_num << endl;Compute();}return 0;}