poj 1088 滑雪 动态规划

题意：n*m矩阵，从某一点起，向上下左右四个方向走，只能走到比自身值小的格子上，问最长路径的长度。

分析：从小到大考虑每一个格子，dp[i][j]=max(dp[i-1][j], dp[i+1][j], dp[i][j+1], dp[i][j-1]) + 1, 其中 a[i][j]>a[x][y]。

 

 

int dx[] = {-1,0,1,0};//up Right down Left
int dy[] = {0,1,0,-1};

const int M = 105;
int a[M][M], d[M][M];
int n, m, ans;

struct node{
    int x,y;
}nd[M*M];
int p;

bool cmp(node n1, node n2){
    return a[n1.x][n1.y]<a[n2.x][n2.y];
}

int main(){
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    #endif

    scanf("%d%d", &n, &m);
    FOR(i, 0, n) FOR(j, 0, m){
        scanf("%d", &a[i][j]);
        nd[p].x=i; nd[p++].y=j;
        d[i][j]=1;
    }
    sort(nd, nd+p, cmp);

    ans = 1;
    FOR(k, 0, p){
        int i=nd[k].x, j=nd[k].y;
        FOR(r, 0, 4){
            int ii=i+dx[r], jj=j+dy[r];
            if(ii<0 || ii>=n || jj<0 || jj>=m) continue;
            if(a[i][j] <= a[ii][jj] || d[i][j] >= d[ii][jj]+1) continue;
            d[i][j] = d[ii][jj]+1;
        }
        checkmax(ans, d[i][j]);
    }

    printf("%d\n", ans);

    return 0;
}