codeforce gym 101726 problem A. Popularity on Facebook

A. Popularity on Facebook
time limit per test
2.0 s
memory limit per test
256 MB
input
standard input
output
standard output

Nowadays everyone is connected, has a Facebook page, posts photos to Instagram and videos to Youtube, and so on. Even GPS systems now are based on social networks, making everything more fun (and harder to understand, but that's another problem).

Being popular on Facebook is almost a need. A person with less than 700, 800 friends may be considered almost a pariah in this new reality.

Maybe that's why some people tend to exaggerate when telling the number of friends they have. Consider a community with N people and, for each one of them, consider we know the number of friends that person says he or she has on this community. Your task is to determine if in fact it is possible that everyone is telling the truth. Remember someone cannot be his or her own friend, and two people cannot be friends more than once.

Input

On the first line T, the number of test cases.

The first line of each test case has an integer N. The second line has N space-separated integers a1, ..., aN. The i-th person claims to have ai friends on the community.

Limits

  • 1 ≤ N ≤ 105
  • 0 ≤ ai ≤ 105
  • The sum of N over all test cases does not exceed 5·105
Output

For each case, print on a single line Y if it is possible that no one is lying, or N otherwise.

Example
input
Copy
2
3
1 1 1
3
2 2 2
output
Copy
N
Y
思路：将度数最小的连接最大的那x个，于是需要一个区间修改的数据结构。同时你删完一段区间后还要保证整体区间的单调性，而更改后不单调递增的话只可能是前面比后面一段大1，故二分得出两端区间左右端点，再进行两次区间修改就行了。
复杂度nlog2n.
  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 int n, t, a[100015];
 45 int val[400015], lazy[400015];
 46 void pushdown(int i) {
 47     if (lazy[i]) {
 48         lazy[i << 1] += lazy[i];
 49         lazy[i << 1 | 1] += lazy[i];
 50         lazy[i] = 0;
 51     }
 52 }
 53 void build(int i, int l, int r) {
 54     lazy[i] = 0;
 55     if (l == r) {
 56         val[i] = a[l];
 57         return;
 58     }
 59     int mi = l + r >> 1;
 60     build(i << 1, l, mi);
 61     build(i << 1 | 1, mi + 1, r);
 62 }
 63 void update(int i, int l, int r, int x, int y, int z) {
 64     if (y < x) return;
 65     if (x <= l && r <= y) {
 66         lazy[i] += z;
 67         return;
 68     }
 69     pushdown(i);
 70     int mi = l + r >> 1;
 71     if (x <= mi) update(i << 1, l, mi, x, y, z);
 72     if (mi < y) update(i << 1 | 1, mi + 1, r, x, y, z);
 73 }
 74 int query(int i, int l, int r, int x) {
 75     if (l == x && x == r) {
 76         return val[i] + lazy[i];
 77     }
 78     int mi = l + r >> 1;
 79     pushdown(i);
 80     if (x <= mi) return query(i << 1, l, mi, x);
 81     if (mi < x) return query(i << 1 | 1, mi + 1, r, x);
 82 }
 83 int main() {
 84     scanf("%d", &t);
 85     while (t--) {
 86         scanf("%d", &n);
 87         for (int i = 1; i <= n; ++i) {
 88             scanf("%d", &a[i]);
 89         }
 90         sort(a + 1, a + n + 1);
 91         build(1, 1, n);
 92         int flag = 1;
 93         for (int i = 1; i < n; ++i) {
 94             int x = query(1, 1, n, i), p1 = n - x + 1;
 95             if (!x) continue;
 96             if (p1 <= i) {
 97                 flag = 0;
 98                 break;
 99             }
100             update(1, 1, n, p1, n, -1);
101             int v1 = query(1, 1, n, p1);
102             if (p1 < 0) {
103                 flag = 0;
104                 break;
105             }
106             int s = i + 1, e = p1 - 1, mi, p2 = p1;
107             while (s <= e) {
108                 mi = s + e >> 1;
109                 if (v1 < query(1, 1, n, mi)) {
110                     e = (p2 = mi) - 1;
111                 } else {
112                     s = mi + 1;
113                 }
114             }
115             s = p1 + 1, e = n;
116             int p3 = p1;
117             while (s <= e) {
118                 mi = s + e >> 1;
119                 if (v1 == query(1, 1, n, mi)) {
120                     s = (p3 = mi) + 1;
121                 } else {
122                     e = mi - 1;
123                 }
124             }
125             int l1 = p1 - p2, l2 = p3 - p1 + 1;
126             if (l1 < l2) {
127                 update(1, 1, n, p2, p2 + l1 - 1, -1);
128                 update(1, 1, n, p3 - l1 + 1, p3, 1);
129             } else {
130                 update(1, 1, n, p2, p2 + l2 - 1, -1);
131                 update(1, 1, n, p3 - l2 + 1, p3, 1);
132             }
133         }
134         if (query(1, 1, n, n)) flag = 0;
135         puts(flag ? "Y" : "N");
136     }
137     return 0;
138 }
View Code

 

posted @ 2018-04-28 19:36  hit_yjl  阅读(357)  评论(0)  编辑  收藏  举报