poj1185 炮兵阵地【状压DP】

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <algorithm>
  4 #include <cmath>
  5 #include <cstring>
  6 
  7 using namespace std;
  8 typedef long long LL;
  9 #define inf 0x3f3f3f3f
 10 
 11 const int maxn = 105;
 12 int n, m, mostm;
 13 int dp[maxn][80][80], dixing[maxn];
 14 int cntone[80], S[80], cnt;
 15 
 16 bool valid(int i, int x)
 17 {
 18     if (dixing[i] & x)return false;
 19     else return true;
 20 }
 21 
 22 int get_one(int x)
 23 {
 24     int res = 0;
 25     while (x) {
 26         res += (x & 1);
 27         x >>= 1;
 28     }
 29     return res;
 30 }
 31 
 32 bool ok(int x)
 33 {
 34     if (x & (x << 1))return false;
 35     if (x & (x << 2)) return false;
 36     return true;
 37 }
 38 
 39 void getS()
 40 {
 41     cnt = 0;
 42     mostm = 1 << m;
 43     for (int i = 0; i < mostm; i++) {
 44         if (ok(i)) {
 45             S[cnt] = i;
 46             cntone[cnt] = get_one(i);
 47             cnt++;
 48         }
 49     }
 50 }
 51 
 52 int main()
 53 {
 54     while (scanf("%d%d", &n, &m) != EOF) {
 55         getS();
 56         memset(dixing, 0, sizeof(dixing));
 57         for (int i = 1; i <= n; i++) {
 58             char tmp[20];
 59             scanf("%s", tmp);
 60             //printf("%s", tmp + 1);
 61             for (int j = 0; j < m; j++) {
 62                 //cout<<tmp[j]<<endl;
 63                 if (tmp[j] == ‘H‘) {
 64                     dixing[i] |= (1 << (m - 1 - j));
 65                 }
 66             }
 67         }
 68 
 69         memset(dp, -1, sizeof(dp));
 70         dp[0][0][0] = 0;
 71         int ans = 0;
 72         for (int j = 0; j < cnt; j++) {
 73             if (valid(1, S[j])) {
 74                 dp[1][j][0] = cntone[j];
 75                 ans = max(ans, dp[1][j][0]);
 76             }
 77         }
 78         //cout<<1<<" "<<ans<<endl;
 79 
 80         for (int i = 2; i <= n; i++) {
 81             int tiaoshi = -1;
 82             for (int j = 0; j < cnt; j++) {
 83                 if (valid(i, S[j])) {
 84                     for (int k = 0; k < cnt; k++) {
 85                         if (valid(i - 1, S[k]) && (S[j] & S[k]) == 0) {
 86                             int mmm = -1;
 87                             for (int l = 0; l < cnt; l++) {
 88                                 if (valid(i - 2, S[l]) && dp[i - 1][k][l] != -1 && (S[j] & S[l]) == 0) {
 89                                     mmm = max(mmm, dp[i - 1][k][l]);
 90                                 }
 91                             }
 92                             //cout<<i - 1<<" "<<mmm<<endl;
 93                             dp[i][j][k] = max(dp[i][j][k], mmm + cntone[j]);
 94                             tiaoshi = max(tiaoshi, dp[i][j][k]);
 95                             if (i == n)ans = max(ans, dp[i][j][k]);
 96                         }
 97                     }
 98                 }
 99             }
100             //cout<<i<<" "<<tiaoshi<<endl;
101         }
102         printf("%d\n", ans);
103     }
104 
105     //return 0;
106 }