POJ 3254 Corn Fields

记录状态的DP.

每一行的状态有2^12种可能（其中一部分状态是无效的），且与前一行的状态相关，

题目要求计算的可以理解为到达最后一行各个有效状态的次数之和。

输入1 for fertile, 0 for infertile，为了计算方便进行取反，0 for fertile, 1 for infertile。

/*Corn Fields Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 2553 Accepted: 1303 Description Farmer John has 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 are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two 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 how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant. Input Line 1: Two space-separated integers: M and N Lines 2..M+1: Line i+1 describes row i of the pasture with N space-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 3 1 1 1 0 1 0 Sample Output 9 Hint Number the squares as follows: 1 2 3 4 There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9. */ #include <stdio.h> #include "string.h" #define MAX_SQUARE_PARCEL_NUM 12 #define MODULO 100000000 #define MAX_STATE_NUM 4096/*2^12*/ int gaiPasture[MAX_SQUARE_PARCEL_NUM]; int gaiStateReachedNum[MAX_SQUARE_PARCEL_NUM][MAX_STATE_NUM]; int CornFieldsmain(void) { int iM; int iN; int iLoopM; int iLoopN; int iTemp; int iStateNum; int iLoop; int iLoopState; int iLoopPreLineState; int iResult = 0; memset(gaiPasture,0,MAX_SQUARE_PARCEL_NUM*sizeof(int)); memset(gaiStateReachedNum,0,MAX_STATE_NUM*MAX_SQUARE_PARCEL_NUM*sizeof(int)); scanf("%d %d",&iM,&iN); iStateNum = (1<<iN); for (iLoopM = 0; iLoopM < iM; iLoopM++) { for (iLoopN = 0; iLoopN < iN; iLoopN++) { scanf("%d",&iTemp); gaiPasture[iLoopM] = gaiPasture[iLoopM]<<1; gaiPasture[iLoopM] |= iTemp; } gaiPasture[iLoopM] = (~gaiPasture[iLoopM])&((iStateNum-1)); } /*first line*/ for (iLoopState = 0; iLoopState < iStateNum; iLoopState++) { /*infertile*/ if (0 != (gaiPasture[0]&iLoopState)) { continue; } /*当前行没有相邻*/ iTemp = 3;//11 for (iLoop = 0; iLoop < iN-1; iLoop++) { if(iTemp == (iTemp&iLoopState)) { break; } iTemp = iTemp<<1; } if (iLoop < iN-1) { continue; } gaiStateReachedNum[0][iLoopState] ++; } for (iLoopM = 1; iLoopM < iM; iLoopM++) { for (iLoopState = 0; iLoopState < iStateNum; iLoopState++) { /*infertile*/ if (0 != (gaiPasture[iLoopM]&iLoopState)) { continue; } /*当前行没有相邻*/ iTemp = 3;//11 for (iLoop = 0; iLoop < iN-1; iLoop++) { if(iTemp == (iTemp&iLoopState)) { break; } iTemp = iTemp<<1; } if (iLoop < iN-1) { continue; } /*与前一行不相邻*/ for (iLoopPreLineState = 0; iLoopPreLineState < iStateNum; iLoopPreLineState++) { if (0 != gaiStateReachedNum[iLoopM-1][iLoopPreLineState] &&0 == (iLoopPreLineState&iLoopState)) { gaiStateReachedNum[iLoopM][iLoopState] += gaiStateReachedNum[iLoopM-1][iLoopPreLineState]; } } } } /*求最后一行的和即结果*/ for (iLoopState = 0; iLoopState < iStateNum; iLoopState++) { if (0 != gaiStateReachedNum[iM-1][iLoopState]) { iResult += gaiStateReachedNum[iM-1][iLoopState]; } } printf("%d/n",iResult%MODULO); return 0; }