POJ 1088

福利题，此前做过类似的DP，思路很简单，定义状态\(dv(i, j)\)为以\((i, j)\)为起点的最长长度

#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxr= 105;

int r, c;
int a[maxr][maxr];
int dv[maxr][maxr];
int step[4][2]= {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
bool vis[maxr][maxr];

int DP(const int i, const int j)
{
	if (vis[i][j]){
		return dv[i][j];
	}

	int &v= dv[i][j];
	for (int k= 0; k< 4; ++k){
		int ni= i+step[k][0], nj= j+step[k][1];
		if (ni>= 0 && ni< r && nj>= 0 && nj< c && a[i][j] > a[ni][nj]){
			v= max(v, DP(ni, nj)+1);
		}
	}
	vis[i][j]= 1;

	return v;
}

int main()
{
	scanf("%d %d", &r, &c);

	for (int i= 0; i< r; ++i){
		for (int j= 0; j< c; ++j){
			scanf("%d", a[i]+j);
			dv[i][j]= 1;
		}
	}
	memset(vis, 0, sizeof(vis));
	int ans= 0;

	for (int i= 0; i< r; ++i){
		for (int j= 0; j< c; ++j){
			ans= max(ans, DP(i, j));
		}
	}
	printf("%d", ans);

	return 0;
}