排序+逆向思维 ACdream 1205 Disappeared Block

 

题目传送门

 

 1 /*  2  从大到小排序，逆向思维，从最后开始考虑，无后向性  3  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--  4  复杂度 O(n + m)，还以为 O(n^2)吓得写了一半就不写了  5 */  6 #include <cstdio>  7 #include <cstring>  8 #include <iostream>  9 #include <algorithm> 10 #include <map> 11 #include <set> 12 #include <string> 13 using namespace std; 14 15 const int MAXN = 1e6 + 10; 16 const int INF = 0x3f3f3f3f; 17 struct Hight 18 { 19 int val, id; 20 }h[MAXN]; 21 int q[MAXN]; 22 int vis[MAXN]; 23 int ans[MAXN]; 24 25 bool cmp(Hight a, Hight b) 26 { 27 return a.val > b.val; 28 } 29 30 int main(void) //ACdream 1205 Disappeared Block 31 { 32 //freopen ("J.in", "r", stdin); 33 34 int t; 35 int n, m; 36 37 scanf ("%d", &t); 38 int cas = 0; 39 while (t--) 40  { 41 memset (vis, 0, sizeof (vis)); 42 scanf ("%d%d", &n, &m); 43 for (int i=1; i<=n; ++i) 44  { 45 scanf ("%d", &h[i].val); 46 h[i].id = i; 47  } 48 for (int i=1; i<=m; ++i) scanf ("%d", &q[i]); 49 50 sort (h+1, h+1+n, cmp); 51 52 int cnt = 0; int res = 0; 53 for (int i=1, j=m; j>=1; --j) 54  { 55 for (; i<=n; ++i) 56  { 57 if (h[i].val <= q[j]) break; 58 int pos = h[i].id; 59 vis[h[i].id] = 1; 60 if (!vis[pos-1] && !vis[pos+1]) res++; 61 if (vis[pos-1] && vis[pos+1]) res--; 62  } 63 ans[j] = res; 64  } 65 printf ("Case #%d: ", ++cas); 66 for (int i=1; i<=m; ++i) printf ("%d%c", ans[i], (i==m) ? '\n' : ' '); 67  } 68 69 70 return 0; 71 } 72 73 /* 74 Case #1: 1 1 2 75 Case #2: 1 2 1 1 0 76 */