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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 83899 | Accepted: 31413 |
Description
1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
Input
Output
Sample Input
5 5 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
Sample Output
25
Source
#include<stdio.h>
#include<string.h>
int dp[110][110];
int a[110][110];
int max(int a,int b){
return a>b?a:b;
}
int n,m;
int s[4][2]={1,0,0,1,-1,0,0,-1};
int dfs(int i,int j){
if(dp[i][j]) return dp[i][j];
for(int k=0;k<4;++k){
if(i+s[k][0]>0&&i+s[k][0]<=n&&j+s[k][1]>0&&j+s[k][1]<=m)
if(a[i][j]>a[i+s[k][0]][j+s[k][1]]){
dp[i][j]=max(dp[i][j],dfs(i+s[k][0],j+s[k][1])+1);
}
}
return dp[i][j];
}
int main(){
while(~scanf("%d%d",&n,&m)){
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
int i,j,res=0;
for(i=1;i<=n;++i){
for(j=1;j<=m;++j){
scanf("%d",&a[i][j]);
}
}
for(i=1;i<=n;++i){
for(j=1;j<=m;++j){
res=max(res,dfs(i,j));
}
}
printf("%d\n",res+1);
}
return 0;
}版权声明:本文为博主原创文章,未经博主允许不得转载。
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原文地址:http://blog.csdn.net/qq_18062811/article/details/47446299