POJ 1185 炮兵阵地 (状压DP)

题意：中文题。

析：dp[i][s][t] 表示第 i 行状态为 s， 第 i-1 行为 t，然后就很简单了，但是要超内存，实际上状态最多才60个，所以后两维开60就好，

然后又超时间，就一直加优化，提前预处理。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 100000000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int dp[105][65][65];
int st[102], sold[1100];
vector<int> al;
int id[1100];
vector<int> state[65];
vector<int> thr[65][65];

bool judge(int s){
  if(m == 1)  return true;
  if((s&1) && (s&2))  return false;
  for(int i = 2; i < m; ++i)
    if((s&(1<<i)) && ((s&(1<<i-1))||(s&(1<<i-2))))  return false;
  return true;
}

int calc(int s){
  int ans = 0;
  for(int i = 0; i < m; ++i)
    if(s & (1<<i))  ++ans;
  return ans;
}

bool judge1(int s, int t){
  for(int i = 0; i < m; ++i)
    if(!(s&(1<<i)) && (t&(1<<i)))  return false;
  return true;
}

bool judge2(int s, int t){
  for(int i = 0; i < m; ++i)
    if(s&(1<<i) && (t&(1<<i)))  return false;
  return true;
}

void solve(vector<int> &v, int s){
  for(int i = 0; i < al.size(); ++i)
    if(judge2(s, al[i]))  v.push_back(al[i]);
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 1; i <= n; ++i){
      char s[15];
      st[i] = 0;
      scanf("%s", s);
      for(int j = 0; j < m; ++j)
        if(s[j] == 'P')  st[i] |= 1<<j;
    }
    int all = 1 << m;
    al.clear();
    int cnt = 0;
    for(int i = 0; i < all; ++i)
      if(judge(i)){
        al.push_back(i);
        sold[i] = calc(i);
        id[i] = cnt++;
      }

    memset(dp, 0, sizeof dp);
    int ans = 0;
    for(int i = 0; i < al.size(); ++i){
      state[i].clear();
      solve(state[i], al[i]);
      if(judge1(st[1], al[i])){
        int idd = id[al[i]];
        dp[1][idd][0] = calc(al[i]);
        ans = max(ans, dp[1][idd][0]);
      }
    }
    for(int i = 0; i < 65; ++i)
      for(int j = 0; j < 65; ++j)
       thr[i][j].clear();
    for(int i = 0; i < al.size(); ++i)
      for(int j = 0; j < state[i].size(); ++j)
        for(int k = 0; k < al.size(); ++k)
          if(judge2(al[i], al[k]) && judge2(state[i][j], al[k])){
            thr[i][id[state[i][j]]].push_back(al[k]);
            thr[id[state[i][j]]][i].push_back(al[k]);
          }

    for(int i = 2; i <= n; ++i)
      for(int ii = 0; ii < al.size(); ++ii){
        int j = al[ii];
        if(!judge1(st[i], j))  continue;
        int idj = id[j];
        for(int k = 0; k < state[idj].size(); ++k){
          int kk = state[idj][k];
          if(!judge1(st[i-1], kk))  continue;
          int idk = id[kk];
          int &res = dp[i][idj][idk];
          for(int l = 0; l < thr[idj][idk].size(); ++l){
            int ll = thr[idj][idk][l];
            int idl = id[ll];
            res = max(res, dp[i-1][idk][idl]);
          }
          res += sold[j];
          ans = max(ans, res);
        }
      }
    printf("%d
", ans);
  }
  return 0;
}