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 = 10000;
const int MAXM = 100005;
const double eps = 1e-13;
//const LL MOD = 1000000007;

int N, M;
int ma[110];
int dp[2][200][200];

int stNum;
int st[1000], num[1000];

int judge(int x)
{
    if(x & (x<<2)) return 0;
    return 1;
}

int get1(int x)
{
    int ret = 0;
    while(x)
    {
        ret += x & 1;
        x >>= 1;
    }
    return ret;
}

void init()
{
    mem1(dp);mem0(ma);
    stNum = 0;
    for(int i=0;i<(1<<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 = 0;
        for(int i=1;i<=N;i++)
            for(int j=0;j<M;j++) {
                scanf("%d", &x);
                if(x==0) ma[i] |= (1<<j);
            }
        for(int i=0;i<stNum;i++) {
            if(st[i] & ma[1]) continue;
            dp[cur][i][0] = num[i];
        }
        for(int r = 2; r <= N; r ++)
        {
            cur = !cur;
            for(int i = 0; i < stNum; i ++)
            {
                if(st[i] & ma[r]) continue;
                for(int j = 0; j < stNum; j ++ )
                {
                    if(st[j] & ma[r-1]) continue;
                    int p = (st[i]<<1) | (st[i]>>1);
                    if(st[j] & p) continue;
                    for(int k = 0; k < stNum; k ++) if(dp[!cur][j][k] != -1)
                    {
                        int pp = st[i] | ((st[j]<<1) | (st[j]>>1));
                        if(st[k] & pp) continue;
                        dp[cur][i][j] = MAX(dp[cur][i][j], dp[!cur][j][k] + num[i]);
                    }

                }
            }
        }
        int ans = 0;
        for(int i=0;i<stNum;i++) for(int j=0;j<stNum;j++)
            ans = MAX(ans, dp[cur][i][j]);
        printf("%d
", ans);
    }
    return 0;
}


/*

6
3
2
12

*/