# 616 nyoj newbie dp, 616 nyojdp

Source: Internet
Author: User

616 nyoj newbie dp, 616 nyojdp
Newbie dp time limit: 1000 MS | memory limit: 65535 KB difficulty: 3

Description

This is a simple survival game where you control a robot from the starting point () of a board to the ending point (n, m) of the Board ). The rules of the game are described as follows:
1. The robot starts at the starting point of the Board and has the energy marked by the starting point.
2. robots can only go right or down, and each step consumes one unit of energy.
3. Robots cannot stay in place.
4. When the robot selects a feasible path, when he reaches the end of the path, he will only have the energy marked by the end point.

For example, a robot starts at () and has 4 units of energy. The blue square indicates the point he can reach. If the end point he chooses in this path is)

Point. When it reaches (2, 4), it will have 1 unit of energy. This is a way to start the next path selection until it reaches (6, 6.
The question is how many ways robots can go from the start point to the end point. This may be a large number. The output result is a modulo of 10000.

Input
Enter an integer T in the first line to indicate the number of data groups.
For each group of data, enter two integers n, m (1 <= n, m <= 100) in the first row ). Indicates the size of the Board. Next, enter n rows, with m integers e (0 <= e <20) in each row ).
Output
Result of Modulo 10000 for the total number of data output modes of each group.
Sample Input
`16 64 5 6 6 4 32 2 3 1 7 21 1 4 6 2 75 8 4 3 9 57 6 6 2 1 53 1 1 3 7 2`
Sample output
`3948`
Source
Hdu

Liu Cheng

Train of Thought: simulate a robot's walk with a few energy steps and only simulate the path it may walk.

Define dp [I] [j] as the number of conditions to map [I] [j]

Dp [0] [0] = 1;

When map [I] [j] is 0

If (map [I] [j])

For (k = 0; k <= map [I] [j]; k ++)

For (h = 0; h <= map [I] [j]-k; h ++)

If (k! = 0 | h! = 0) & k + I <n & h + j <m) // bottom right walking

Dp [I + k] [h + j] + = dp [I] [j];

`#include<iostream>#include<algorithm>#include<string.h>#include<cstdio>#include<cmath>using namespace std;int dp[105][105],map[105][105];int main(){    int t;    cin>>t; while(t--) {     int n,m;     cin>>n>>m;     for(int i=0;i<n;i++)     for(int j=0;j<m;j++)         cin>>map[i][j];        memset(dp,0,sizeof(dp));    dp[0][0]=1;     for(int i=0;i<n;i++)     for(int j=0;j<m;j++)        if(map[i][j])        for(int k=0;k<=map[i][j];k++)        for(int h=0;h<=map[i][j]-k;h++)        {            if((k!=0||h!=0)&&k+i<n&&h+j<m)            dp[i+k][j+h]=(dp[i][j]+dp[i+k][j+h])%10000;        }    cout<<dp[n-1][m-1]<<endl; }}`

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