POJ 1088

记忆化搜索啊

或者直接对高度排序啊，这样就会先更新步数少的，后更新步数多的了，就可以直接利用先前已经得到的值了

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define max(x, y) (x > y ? x : y)
#define min(x, y) (x > y ? y : x)
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define Yes printf("Yes
")
#define No printf("No
")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};

int h, w;
const int maxn = 1e2 + 10;

int hei[maxn][maxn];
int ans[maxn][maxn];

struct node {
    int x, y, h;
}map[maxn * maxn];


bool cmp(const node& A, const node& B) {
    return A.h < B.h;
}

int main(int argc, const char * argv[]) {
    int cnt = 0;
    scanf("%d%d", &h, &w);
    for (int i = 1; i <= h; i++) {
        for (int j = 1; j <= w; j++) {
            scanf("%d", &hei[i][j]);
            map[cnt].x = i, map[cnt].y = j, map[cnt].h = hei[i][j];
            cnt++;
            ans[i][j] = 1;
        }
    }
    sort(map, map  + cnt, cmp);
    for (int i = 0; i < cnt; i++) {
        int x = map[i].x, y = map[i].y;
        for (int j = 0; j < 4; j++) {
            int nx = x + dx[j], ny = y + dy[j];
            if (hei[x][y] > hei[nx][ny]) {
                ans[x][y] = max(ans[x][y], ans[nx][ny] + 1);
            }
        }
    }
    int res = 0;
    for (int i = 1; i <= h; i++) {
        for (int j = 1; j <= w; j++) {
            res = max(res, ans[i][j]);
        }
    }
    printf("%d
", res);
    return 0;
}