HDU 4539郑厂长系列故事――排兵布阵（状压DP）

HDU 4539  郑厂长系列故事――排兵布阵

基础的状压DP，首先记录先每一行可取的所哟状态（一行里互不冲突的大概160个状态），

直接套了一个4重循环居然没超时我就呵呵了

 //#pragma comment(linker,"/STACK:102400000,102400000")
 #include <map>
 #include <set>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <vector>
 #include <cstdio>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 #define INF 1e8
 #define inf (-((LL)1<<40))
 #define lson k<<1, L, mid
 #define rson k<<1|1, mid+1, R
 #define mem0(a) memset(a,0,sizeof(a))
 #define mem1(a) memset(a,-1,sizeof(a))
 #define mem(a, b) memset(a, b, sizeof(a))
 #define FOPENIN(IN) freopen(IN, "r", stdin)
 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }

 //typedef __int64 LL;
 //typedef long long LL;
 const int MAXN = ;
 const int MAXM = ;
 const double eps = 1e-;
 //const LL MOD = 1000000007;

 int N, M;
 int ma[];
 int dp[][][];

 int stNum;
 int st[], num[];

 int judge(int x)
 {
     if(x & (x<<)) return ;
     return ;
 }

 int get1(int x)
 {
     int ret = ;
     while(x)
     {
         ret += x & ;
         x >>= ;
     }
     return ret;
 }

 void init()
 {
     mem1(dp);mem0(ma);
     stNum = ;
     for(int i=;i<(<<M);i++) if(judge(i))
     {
         st[stNum] = i;
         num[stNum++] = get1(i);
     }
 }

 int main()
 {
     //FOPENIN("in.txt");
     while(~scanf("%d %d%*c", &N, &M))
     {
         init();
         int x, cur = ;
         for(int i=;i<=N;i++)
             for(int j=;j<M;j++) {
                 scanf("%d", &x);
                 if(x==) ma[i] |= (<<j);
             }
         for(int i=;i<stNum;i++) {
             if(st[i] & ma[]) continue;
             dp[cur][i][] = num[i];
         }
         for(int r = ; r <= N; r ++)
         {
             cur = !cur;
             for(int i = ; i < stNum; i ++)
             {
                 if(st[i] & ma[r]) continue;
                 for(int j = ; j < stNum; j ++ )
                 {
                     if(st[j] & ma[r-]) continue;
                     int p = (st[i]<<) | (st[i]>>);
                     if(st[j] & p) continue;
                     for(int k = ; k < stNum; k ++) if(dp[!cur][j][k] != -)
                     {
                         int pp = st[i] | ((st[j]<<) | (st[j]>>));
                         if(st[k] & pp) continue;
                         dp[cur][i][j] = MAX(dp[cur][i][j], dp[!cur][j][k] + num[i]);
                     }

                 }
             }
         }
         int ans = ;
         for(int i=;i<stNum;i++) for(int j=;j<stNum;j++)
             ans = MAX(ans, dp[cur][i][j]);
         printf("%d\n", ans);
     }
     return ;
 }

 /*

 6
 3
 2
 12

 */