POJ 1185炮兵阵地

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解题思路：

简单的状压DP,1表示放炮，预处理出每一行所有两个1间隔不小于2的状态，每一行的状态只和上面两行有关，因此可以枚举这三行的状态，用DP[i][j][k]表示第i行状态为k，第i-1行状态为j的数目，转移方程为dp[i][j][k] = max(dp[i][j][k], dp[i-1][l][j] +count(k));

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>#include <queue>#include <algorithm>#include <set>#include <map>#define LL long long#define FOR(i, x, y) for(int i=x;i<=y;i++)using namespace std;const int MAXNr = 100 + 10;const int MAXNc = (1 << 11);int dp[MAXNr][MAXNr][MAXNr];int C[MAXNr], state[MAXNc];char str[MAXNr][MAXNr];int Map[MAXNr];int N, M, K;bool judge(int x){    if(x & (x << 1)) return false;    if(x & (x << 2)) return false;    return true;}int count(int x){    int res = 0;    while(x)    {        if(x & 1) res++;        x >>= 1;    }    return res;}void init(){    memset(Map, 0, sizeof(Map));    int r = (1 << M) - 1; K = 0;    FOR(i, 0, r) if(judge(i))    state[++K] = i;    memset(dp, -1, sizeof(dp));}int main(){    while(scanf("%d%d", &N, &M)!=EOF)    {        if(N == 0 && M == 0) break;        init();        FOR(i, 1, N) scanf("%s", str[i]+1);        FOR(i, 1, N)        FOR(j, 1, M) if(str[i][j] == 'H')        Map[i] = (Map[i] | (1 << (M - j)));        FOR(i, 1, K) if(!(state[i] & Map[1]))        {            dp[1][1][i] = count(state[i]);        }        FOR(i, 2, N)        {            FOR(l, 1, K)            {                if(state[l] & Map[i]) continue;                FOR(j, 1, K)                {                    if(state[l] & state[j]) continue;                    FOR(t, 1, K)                    {                        if(state[l] & state[t]) continue;                        if(dp[i-1][t][j] == -1) continue;                        dp[i][j][l] = max(dp[i][j][l], dp[i-1][t][j] + count(state[l]));                    }                }            }        }        int ans = 0;        FOR(i, 1, N)        FOR(j, 1, K)        FOR(k, 1, K)        ans = max(ans , dp[N][j][k]);        printf("%d\n", ans);    }    return 0;}

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