HDU 2167 pebbles (State compression DP)

The entry-level status compresses the DP question.

Assume that the number in the square is not adjacent to the position (including the diagonal line), and the sum of the largest is obtained.

Practice: compress the status of each row to the binary number, and set the status DP [I] [p] to the Set P for row I, it is easy to get DP [I] [p] = max {DP [I-1] [Q] + CNT [I] [p] | set p, q can coexist }, CNT [I] [p] is the sum of the numbers obtained when the I Behavior Set P.

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  1 /*  2  *Author:       Zhaofa Fang  3  *Created time: 2013-03-31-10.33  4  *Language:     C++  5  */  6 #include <cstdio>  7 #include <cstdlib>  8 #include <sstream>  9 #include <iostream> 10 #include <cmath> 11 #include <cstring> 12 #include <algorithm> 13 #include <string> 14 #include <utility> 15 #include <vector> 16 #include <queue> 17 #include <stack> 18 #include <map> 19 #include <set> 20 using namespace std; 21  22 typedef long long ll; 23 #define DEBUG(x) cout<< #x << ':' << x << endl 24 #define REP(i,n) for(int i=0;i < (n);i++) 25 #define REPD(i,n) for(int i=(n-1);i >= 0;i--) 26 #define FOR(i,s,t) for(int i = (s);i <= (t);i++) 27 #define FORD(i,s,t) for(int i = (s);i >= (t);i--) 28 #define PII pair<int,int> 29 #define PB push_back 30 #define MP make_pair 31 #define ft first 32 #define sd second 33 #define lowbit(x) (x&(-x)) 34 #define INF (1<<30) 35  36 int k; 37 int maz[17][17],dp[2][1600]; 38 int cnt[20][1600]; 39 int s[1600]; 40 bool ok(int x){ 41     if(x&(x<<1))return false; 42     return true; 43 } 44 int fun(int x,int row,int n){ 45     int res=0,j=0; 46     REP(j,n){ 47         if(x&(1<<j))res+=maz[row][j]; 48     } 49     return res; 50 } 51 void pre_deal(int n){ 52  53     memset(s,0,sizeof(s)); 54     memset(cnt,0,sizeof(cnt)); 55     REP(i,(1<<n)){ 56         if(ok(i)){ 57             s[k++] = i; 58         } 59     } 60     REP(i,n){ 61         REP(j,k) 62         cnt[i][j] = fun(s[j],i,n); 63     } 64 } 65  66 int main(){ 67     //freopen("in","r",stdin); 68     //freopen("out","w",stdout); 69     char str[100]; 70     while(gets(str)){ 71         int len = strlen(str); 72         int n=0,res=0; 73         REP(i,len){ 74             if(str[i]>='0'&&str[i]<='9')res = res*10+str[i]-'0'; 75             else { 76                 maz[0][n++] = res; 77                 res = 0; 78             } 79         } 80         maz[0][n++]=res; 81         FOR(i,1,n-1)REP(j,n)scanf("%d",&maz[i][j]); 82         getchar(); 83         gets(str); 84         k = 0; 85         pre_deal(n); 86         memset(dp,-1,sizeof(dp)); 87         int now = 1; 88         REP(i,k)dp[now][i] = cnt[0][i]; 89         FOR(i,1,n-1){ 90             now ^= 1; 91             REP(p,k){ 92                 REP(q,k){ 93                     if(s[p]&s[q])continue; 94                     if(s[p]&(s[q]<<1))continue; 95                     if(s[p]&(s[q]>>1))continue; 96                     if(dp[now^1][q]==-1)continue; 97                     dp[now][p] = max(dp[now][p],dp[now^1][q]+cnt[i][p]); 98                 } 99             }100         }101         int ans = 0;102         REP(i,k)ans=max(ans,dp[now][i]);103         printf("%d\n",ans);104     }105     return 0;106 }