HDU 2151 Worm

题目是中文,题意就不啰嗦了.

设dp[i][j] 为第i分钟后到达第j颗树的方案数.

dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j + 1] 满足j - 1>= 0 && j + 1 <= N && dp[i - 1][j - 1] != -1 && dp[i - 1][j + 1] != -1

初始化为dp为-1

base case: dp[0][P] = 1

#include <cstdio>
#include <memory.h>
using namespace std;
const int MAX = 101;

int dp[MAX][MAX];
int main(int argc, char const *argv[]){
int N, P, M, T;
while(~scanf("%d%d%d%d", &N, &P, &M, &T)){
memset(dp, -1, sizeof(dp));
dp[0][P] = 1;
for(int i = 0; i < M; ++i){
for(int j = 1; j <= N; ++j){
if(dp[i][j] != -1){
if(j - 1 >= 1){
if(dp[i + 1][j - 1] == -1){
dp[i + 1][j - 1] = dp[i][j];
}else{
dp[i + 1][j - 1] += dp[i][j];
}
}
if(j + 1 <= N){
if(dp[i + 1][j + 1] == -1){
dp[i + 1][j + 1] = dp[i][j];
}else{
dp[i + 1][j + 1] += dp[i][j];
}
}
}
}
}
if(dp[M][T] == -1){
printf("0\n");
}else{
printf("%d\n", dp[M][T]);
}
}
return 0;
}