hdu1978基础dp

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Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
思路：简单dp，从点(x,y)可以走到的点的表达式是(dx + x,dy + y) && x + y <= a[x][y] && x + dx <= n && y + dy <= m
此时就可以进行状态的转移了dp[x + dx][y + dy] = (dp[x][y] + dp[x + dx][y + dy]) % 10000;
最后的结果状态在dp[n][m]中；
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// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
#define pk push_back
template<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}
template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}
typedef long long ll;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dp[110][110];
int a[110][110];
int main()
{	
	// ios::sync_with_stdio(false);
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int n,m;
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&m);
		for (int i = 1;i <= n;++i)
			for (int j = 1;j <= m;++j)
				scanf("%d",&a[i][j]);
		MEM(dp,0);
		dp[1][1] = 1;
		for (int i = 1;i <= n;++i){
			for (int j = 1;j <= m;++j){
				for (int dx = 0;dx <= a[i][j];dx++){
					if (i + dx > n) break;
					for (int dy = 0;dy + dx <= a[i][j];dy++){
						if (dx == 0 && dy == 0) continue;
						if (j + dy > m) break;
						dp[i + dx][j + dy] = (dp[i][j] + dp[i + dx][j + dy]) % 10000;
					}
				}
			}
		}
		cout << dp[n][m] << endl;
	}
	return 0;
}