poj-3254 Corn fields (pressure DP)

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Corn fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12175		Accepted: 6398

Description

Farmer John had purchased a lush new rectangular pasture composed of M by N (1≤ M ≤ 12; 1≤ n ≤12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares is infertile and can ' t be planted. Canny FJ knows that the cows dislike eating close to all other, so when choosing which squares to plant, he avoids choosi Ng squares that is adjacent; No. Chosen squares share an edge. He has not yet made the final choice as to which squares to plant.

Being a very open-minded man, Farmer John wants to consider all possible options for what to choose the squares for Plantin G. He's so open-minded that he considers choosing no squares as a valid option! Farmer John determine the number of ways he can choose the squares to plant.

Input

Line 1:two space-separated integers: Mand N
Lines 2.. M+1:line I+1 describes row Iof the pasture with Nspace-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)

Output

Line 1:one integer:the number of ways that FJ can choose the squares modulo 100,000,000.

Sample Input

2 31 1 10 1 0

Sample Output

9

Test instructions

To N*m lattice, the adjacent lattice can not all plant, now give 0 of the lattice can not plant, ask you how many kinds of plant methods;

Ideas:

DP[I][J] indicates that the status of line I is J before I line a total number of methods, and then is the legal state to find and transfer, the specific look at the code;

AC Code:

//#include <bits/stdc++.h>#include <vector>#include<iostream>#include<queue>#include<cmath>#include<map>#include<cstring>#include<algorithm>#include<cstdio>using namespacestd;#defineRiep (n) for (int i=1;i<=n;i++)#defineRIOP (n) for (int i=0;i<n;i++)#defineRJEP (n) for (int j=1;j<=n;j++)#defineRJOP (n) for (int j=0;j<n;j++)#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong Longll;template<classT>voidRead (t&num) {    CharCH;BOOLf=false;  for(Ch=getchar (); ch<'0'|| Ch>'9'; f= ch=='-', ch=GetChar ());  for(num=0; ch>='0'&&ch<='9'; num=num*Ten+ch-'0', ch=GetChar ()); F&& (num=-num);}intstk[ -], Tp;template<classT> Inlinevoidprint (T p) {if(!p) {Puts ("0");return; }  while(p) stk[++ TP] = p%Ten, p/=Ten;  while(TP) Putchar (stk[tp--] +'0'); Putchar ('\ n');}ConstLL mod=1e8;Const DoublePi=acos (-1.0);ConstLL inf=1e18;Const intN= (1<< A)+Ten;Const intmaxn=1005; LL dp[ -][n];inta[ the][ the],temp[ the],num[ the],cnt,b[n];intn,m;intOkintXinty) {    if(x&y)return 0; return 1;}intCheckintx) {MST (temp,0); ints=0;  while(x) {temp[++s]= (x&1); X>>=1; }     for(intI=1; i<=s;i++)    {        if(temp[i]&&temp[i+1])return 0; }    return 1;}voidInit () {CNT=0;  for(intI=0;i< (1&LT;&LT;M); + +i) {if(check (i)) b[++cnt]=i; }}intMain () {read (n); read (m);  for(intI=1; i<=n;i++)         for(intj=1; j<=m;j++) read (a[i][j]);  for(intI=1; i<=n;i++)    {        intx=0, w=1;  for(intj=1; j<=m;j++)        {            if(!a[i][j]) x+=W; W*=2; } Num[i]=x;    } Init ();  for(intI=1; i<=cnt;i++)    {        if(OK (num[1],b[i]) dp[1][i]=1;//Ans+=dp[1][i];    }     for(intI=2; i<=n;i++)    {         for(intj=1; j<=cnt;j++)        {            if(OK (Num[i],b[j])) { for(intk=1; k<=cnt;k++)                if(OK (num[i-1],B[K]) &&OK (B[j],b[k])) Dp[i][j]+=dp[i-1][k],dp[i][j]%=mod;//Ans+=dp[i][j],ans%=mod;}}} LL ans=0;  for(intj=1; j<=cnt;j++)            if(OK (num[n],b[j])) ans+=dp[n][j],ans%=MoD; cout<<ans<<"\ n"; return 0;}

poj-3254 Corn fields (pressure DP)