Poj 2411 new statement

Today I started srm, and PT thought of dp but I couldn't help it. But after reading rng_58's God code, I felt like a sea of Sky (staring at one afternoon, I can't bear to look straight,

TC is really a good place... Like!

In fact, it is to write the ordinary brick-laying problem in a way similar to the plug dp by Lattice recurrence. The following code should be understandable.

#include <cstdio>  #include <cstring>  #include <algorithm>  long long dp[2][1<<11]; int main() {     int n , m;     while(scanf("%d%d",&n,&m),n||m) {       int cur = 0 , nxt = 1;       dp[cur][0] =  1;        for(int i = 0; i < n; i++) {           for(int j = 0; j < m; j++) {             memset(dp[nxt],0,sizeof(dp[nxt]));             for(int s = 0; s < (1<<m); s++) if(dp[cur][s]){                 if(s&1){dp[nxt][s>>1] += dp[cur][s];continue;}                 if(j+1<m && !(s&2) ){                     int mask = ( s | 2 ) >> 1;                     dp[nxt][mask] += dp[cur][s];                 }                 if(i+1<n) {                     int mask = (s | (1<<m)) >> 1;                     dp[nxt][mask] += dp[cur][s];                 }             }             std::swap(cur,nxt);           }        }        printf("%I64d\n",dp[cur][0]);     }     return 0; } #include <cstdio>#include <cstring>#include <algorithm>long long dp[2][1<<11];int main() {    int n , m;    while(scanf("%d%d",&n,&m),n||m) {      int cur = 0 , nxt = 1;      dp[cur][0] =  1;       for(int i = 0; i < n; i++) {          for(int j = 0; j < m; j++) {            memset(dp[nxt],0,sizeof(dp[nxt]));            for(int s = 0; s < (1<<m); s++) if(dp[cur][s]){                if(s&1){dp[nxt][s>>1] += dp[cur][s];continue;}                if(j+1<m && !(s&2) ){                    int mask = ( s | 2 ) >> 1;                    dp[nxt][mask] += dp[cur][s];                }                if(i+1<n) {                    int mask = (s | (1<<m)) >> 1;                    dp[nxt][mask] += dp[cur][s];                }            }            std::swap(cur,nxt);          }       }       printf("%I64d\n",dp[cur][0]);    }    return 0;}

The following is the example of a year ago...

 

#include<stdio.h>  #include<string.h>  int h,w; __int64 dp[15][2050]; void dfs1(int row,int state,int col,int state2){     if(col>w) return ;     if(col==w) {         dp[row+1][state2]+=dp[row][state];         return ;     }     if(!(state&(1<<col))) dfs1(row,state,col+1,state2+(1<<col));     else dfs1(row,state,col+1,state2); } void dfs2(int row,int state,int col,int state2){     if(col>w) return ;     if(col==w) {         if(state2!=state)              dp[row][state2]+=dp[row][state];         return ;     }     if(!(state&(1<<col)) &&   !(state&(1<<(col+1))))         dfs2(row,state,col+2,state2+(1<<col) + (1<<(col+1)));     dfs2(row,state,col+1,state2); } void gao(){     int i,j;     dp[0][(1<<w)-1]=1;     for(i=0;i