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

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <vector>
 6 #define max(x, y) (x > y ? x : y)
 7 #define min(x, y) (x > y ? y : x)
 8 #define INF 0x3f3f3f3f
 9 #define MOD 1000000007
10 #define Yes printf("Yes\n")
11 #define No printf("No\n")
12 using namespace std;
13 typedef long long LL;
14 typedef pair<int, int> PII;
15
16 int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
17
18 int h, w;
19 const int maxn = 1e2 + 10;
20
21 int hei[maxn][maxn];
22 int ans[maxn][maxn];
23
24 struct node {
25     int x, y, h;
26 }map[maxn * maxn];
27
28
29 bool cmp(const node& A, const node& B) {
30     return A.h < B.h;
31 }
32
33 int main(int argc, const char * argv[]) {
34     int cnt = 0;
35     scanf("%d%d", &h, &w);
36     for (int i = 1; i <= h; i++) {
37         for (int j = 1; j <= w; j++) {
38             scanf("%d", &hei[i][j]);
39             map[cnt].x = i, map[cnt].y = j, map[cnt].h = hei[i][j];
40             cnt++;
41             ans[i][j] = 1;
42         }
43     }
44     sort(map, map  + cnt, cmp);
45     for (int i = 0; i < cnt; i++) {
46         int x = map[i].x, y = map[i].y;
47         for (int j = 0; j < 4; j++) {
48             int nx = x + dx[j], ny = y + dy[j];
49             if (hei[x][y] > hei[nx][ny]) {
50                 ans[x][y] = max(ans[x][y], ans[nx][ny] + 1);
51             }
52         }
53     }
54     int res = 0;
55     for (int i = 1; i <= h; i++) {
56         for (int j = 1; j <= w; j++) {
57             res = max(res, ans[i][j]);
58         }
59     }
60     printf("%d\n", res);
61     return 0;
62 }