排序+逆向思维 ACdream 1205 Disappeared Block

题目传送门

 /*
     从大到小排序，逆向思维，从最后开始考虑，无后向性
     每找到一个没被淹没的，对它左右的楼层查询是否它是孤立的，若是++，若不是--
     复杂度 O(n + m)，还以为 O(n^2)吓得写了一半就不写了
 */
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <string>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 struct Hight
 {
     int val, id;
 }h[MAXN];
 int q[MAXN];
 int vis[MAXN];
 int ans[MAXN];

 bool cmp(Hight a, Hight b)
 {
     return a.val > b.val;
 }

 int main(void)        //ACdream 1205 Disappeared Block
 {
     //freopen ("J.in", "r", stdin);

     int t;
     int n, m;

     scanf ("%d", &t);
     int cas = ;
     while (t--)
     {
         memset (vis, , sizeof (vis));
         scanf ("%d%d", &n, &m);
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &h[i].val);
             h[i].id = i;
         }
         for (int i=; i<=m; ++i)    scanf ("%d", &q[i]);

         sort (h+, h++n, cmp);

         int cnt = ;    int res = ;
         for (int i=, j=m; j>=; --j)
         {
             for (; i<=n; ++i)
             {
                 if (h[i].val <= q[j])    break;
                 int pos = h[i].id;
                 vis[h[i].id] = ;
                 if (!vis[pos-] && !vis[pos+])    res++;
                 if (vis[pos-] && vis[pos+])    res--;
             }
             ans[j] = res;
         }
         printf ("Case #%d: ", ++cas);
         for (int i=; i<=m; ++i)    printf ("%d%c", ans[i], (i==m) ? '\n' : ' ');
     }

     return ;
 }

 /*
 Case #1: 1 1 2
 Case #2: 1 2 1 1 0
 */