Nyoj_1273_ propaganda Wall

/*
Propaganda wall
Time limit: Ms | Memory Limit: 65535 KB
Difficulty: 4

Describe

The ALPHA town is beautiful, the roads are neat and clean, and there are so many tourists to visit. The CBA mayor is preparing a series of propaganda on a wall 4*n the south of a road.
In order to unify the plan, the CBA mayor asks each propaganda column only to occupy the adjacent two squares position. But the road is divided by another road into about two paragraphs.
CBA mayor want to know, if each location is affixed to the propaganda column, left and right two paragraphs each have how many different posting scheme. For example: N=6,m=3, k=2, left,
There are 5 different posting schemes on the right.

Input
The first line: T means the following have T Group test data (1≤T≤8)
Next there are t-lines, three positive integers per line, N M K for the length of the road, and the beginning and width of the other road.
(1≤n, M≤1 000 000, 1≤k≤100000)
Output
Each set of test data, the output is one line: two integers, representing the left and right two different posting scenarios. As the total number of programmes
May be large, please output the result after modulo 997.
Sample input

2
6 3 2
5 3 2

Sample output

5 5
5 1

*/

State compression DP, very typical of a chestnut. Also want to say, with the scrolling array is better, we open the array almost explode.

#include <stdio.h> #include <string.h> #include <stdlib.h> const int col = 4;
int row;

int dp[900000][1<<4];
		void Dfs (int r,int c,int pre,int now) {if (C==col) {Dp[r][now] + = Dp[r-1][pre];
		Dp[r][now]%= 997;
	return;
		} if (C+1<=col) {DFS (r,c+1,pre<<1, (now<<1) | |);
	DFS (r,c+1, (pre<<1) | |, now<<1);
	} if (C+2<=col) {DFS (r,c+2, (pre<<2) |3, (now<<2) |3);
	}} int main () {int t,n,m,k;
	scanf ("%d", &t);
		while (t--) {scanf ("%d%d%d", &n,&m,&k);
		row = M-1;
		if (row<=0) {printf ("0");
			} else {memset (dp,0,sizeof (DP));
			dp[0][(1<<col)-1] = 1;
			for (int i=1;i<=row;i++) {DFS (i,0,0,0);
		} printf ("%d", dp[row][(1<<col)-1]);
		} row = n-m-K + 1;
		if (row<=0) {printf ("0\n");
			} else {memset (dp,0,sizeof (DP));
			dp[0][(1<<col)-1] = 1;
			for (int i=1;i<=row;i++) {DFS (i,0,0,0); } printf ("%d\n", dp[row][(1<<col)-1]);
}} return 0;
 }