一道dfs和dp结合的好题 --- Longest Run on a SnowboardUVA-10285

#include <iostream>
#include <cstdio>
#include <string.h>

using namespace std;
const int MX = 100+10;
int vis[MX][MX]; //vis数组记忆化搜索，防止重复。但是在这道题貌似不需要。。。
int dp[MX][MX], mp[MX][MX];
int n, m, ans;
string name;
int xx[] = {1, 0, -1, 0};
int yy[] = {0, 1, 0, -1}; //四个方向


int dfs(int x, int y)
{
    if(vis[x][y]) return dp[x][y];
    for(int i = 0; i < 4; ++i)
    {
        int x1 = x+xx[i];
        int y1 = y+yy[i];
        if(x1 >= 1 && x1 <= n && y1 >= 1 && y1 <= m && mp[x1][y1] < mp[x][y]) //注意递归边界
        {
            dp[x][y] = max(dfs(x1, y1)+1, dp[x][y]); //选权值最大的边
        }
    }
    vis[x][y] = 1; //记忆化搜索优化
    return dp[x][y];
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        ans = 0;
        name.clear();
        memset(vis, 0, sizeof(vis));
        memset(dp, 0, sizeof(dp));
        memset(mp, 0, sizeof(mp));
        cin >> name;
        scanf("%d %d", &n, &m);

        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
                scanf("%d", &mp[i][j]); //初始化mp数组

        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
            {
                int h = dfs(i, j);
                ans = max(ans, h);
            }
        cout << name << ": " << ans+1 << endl; //起始点也算
    }
}