HDU 1078 FatMouse and Cheese

/*记忆化搜索P - FatMouse and Cheese题意: 从（0，0）出发，每次最多走k格，但每次只能往值比它现在在的值大的格子走。题解：dp[i][j] 代表走到（i,j）的最大总值。dp[i][j] = max(dp[prei][prej]+mp[i][j]){i,j 是 prei,prej能走到的格子}贡献1T，1Wa，注意题意走K步是只能笔直走K步，导致T。我以为我nextx是不断更新的，没乘j，导致wa。*/#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<map>using namespace std;const int N = 110;int n,k;int dx[5] = {0,0,1,-1};int dy[5] = {1,-1,0,0};int mp[N][N],v[N][N];int dp[N][N];bool pan(int i,int j){    if(i >= 0 && i < n && j >= 0 && j < n)        return true;    return false;}int DFS(int x,int y){    int &ans = dp[x][y];    if(v[x][y])   return ans;    ans = mp[x][y];    v[x][y] = 1;    for(int i = 0; i < 4; i++)    {        for(int j = 1; j <= k; j++)        {            int nextx = x+dx[i]*j,nexty = y+dy[i]*j;            if(pan(nextx,nexty) && mp[x][y] < mp[nextx][nexty])                ans = max(ans,DFS(nextx,nexty)+mp[x][y]);        }    }    return ans;}int main(){    while(cin >> n >> k)    {        if(n == -1 && k == -1)            break;        memset(dp,0,sizeof(dp));        memset(mp,0,sizeof(mp));        memset(v,0,sizeof(v));        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                scanf("%d",&mp[i][j]);        int ans = DFS(0,0);        cout << ans << endl;    }    return 0;}