HDU 4539郑厂长系列故事――排兵布阵（状压DP）
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
1 //#pragma comment(linker,"/STACK:102400000,102400000") 2 #include <map> 3 #include <set> 4 #include <stack> 5 #include <queue> 6 #include <cmath> 7 #include <ctime> 8 #include <vector> 9 #include <cstdio> 10 #include <cctype> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 using namespace std; 16 #define INF 1e8 17 #define inf (-((LL)1<<40)) 18 #define lson k<<1, L, mid 19 #define rson k<<1|1, mid+1, R 20 #define mem0(a) memset(a,0,sizeof(a)) 21 #define mem1(a) memset(a,-1,sizeof(a)) 22 #define mem(a, b) memset(a, b, sizeof(a)) 23 #define FOPENIN(IN) freopen(IN, "r", stdin) 24 #define FOPENOUT(OUT) freopen(OUT, "w", stdout) 25 template<class T> T CMP_MIN(T a, T b) { return a < b; } 26 template<class T> T CMP_MAX(T a, T b) { return a > b; } 27 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 28 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 29 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 30 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 31 32 //typedef __int64 LL; 33 //typedef long long LL; 34 const int MAXN = 10000; 35 const int MAXM = 100005; 36 const double eps = 1e-13; 37 //const LL MOD = 1000000007; 38 39 int N, M; 40 int ma[110]; 41 int dp[2][200][200]; 42 43 int stNum; 44 int st[1000], num[1000]; 45 46 int judge(int x) 47 { 48 if(x & (x<<2)) return 0; 49 return 1; 50 } 51 52 int get1(int x) 53 { 54 int ret = 0; 55 while(x) 56 { 57 ret += x & 1; 58 x >>= 1; 59 } 60 return ret; 61 } 62 63 void init() 64 { 65 mem1(dp);mem0(ma); 66 stNum = 0; 67 for(int i=0;i<(1<<M);i++) if(judge(i)) 68 { 69 st[stNum] = i; 70 num[stNum++] = get1(i); 71 } 72 } 73 74 75 int main() 76 { 77 //FOPENIN("in.txt"); 78 while(~scanf("%d %d%*c", &N, &M)) 79 { 80 init(); 81 int x, cur = 0; 82 for(int i=1;i<=N;i++) 83 for(int j=0;j<M;j++) { 84 scanf("%d", &x); 85 if(x==0) ma[i] |= (1<<j); 86 } 87 for(int i=0;i<stNum;i++) { 88 if(st[i] & ma[1]) continue; 89 dp[cur][i][0] = num[i]; 90 } 91 for(int r = 2; r <= N; r ++) 92 { 93 cur = !cur; 94 for(int i = 0; i < stNum; i ++) 95 { 96 if(st[i] & ma[r]) continue; 97 for(int j = 0; j < stNum; j ++ ) 98 { 99 if(st[j] & ma[r-1]) continue; 100 int p = (st[i]<<1) | (st[i]>>1); 101 if(st[j] & p) continue; 102 for(int k = 0; k < stNum; k ++) if(dp[!cur][j][k] != -1) 103 { 104 int pp = st[i] | ((st[j]<<1) | (st[j]>>1)); 105 if(st[k] & pp) continue; 106 dp[cur][i][j] = MAX(dp[cur][i][j], dp[!cur][j][k] + num[i]); 107 } 108 109 } 110 } 111 } 112 int ans = 0; 113 for(int i=0;i<stNum;i++) for(int j=0;j<stNum;j++) 114 ans = MAX(ans, dp[cur][i][j]); 115 printf("%d\n", ans); 116 } 117 return 0; 118 } 119 120 121 /* 122 123 6 124 3 125 2 126 12 127 128 */
