Educational Codeforces Round 12 D. Simple Subset(最大团)

题意:

$给定N\le 10^3个数,从中选出一些数,使得这些数任意两两之和是素数$
$求最多选出的数的个数,以及方案$

分析:

$讲道理这题可以贪心,素数=奇数+偶数,所以不考虑1这个数的话答案最多是2$
$- - 最大团直接裸搞也可以,迷之复杂度1000个点都O(跑得过)$

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
//
// Created by TaoSama on 2016-04-29
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
struct MaxClique {
static const int V = 1e3 + 10;
bool g[V][V];
int n, ans, max[V], adj[V][V];
int path[V], clique[V]; //for record
//max[i]:= [i, n]'s maximum clique
//adj[dep][i]:= available vertices
void init(int _n) { n = _n; }
bool dfs(int cur, int dep) {
if(cur == 0) {
if(dep > ans) {
ans = dep;
swap(clique, path);
return 1;
}
return 0;
}
for(int i = 0; i < cur; ++i) {
if(dep + cur - i <= ans) return 0; //dep + left <= ans
int u = adj[dep][i], nxt = 0;
if(dep + max[u] <= ans) return 0; //same as above
path[dep] = u;
for(int j = i + 1; j < cur; ++j) {
int v = adj[dep][j];
if(g[u][v]) adj[dep + 1][nxt++] = v;
}
if(dfs(nxt, dep + 1)) return 1;
}
return 0;
}
int maxClique() {
ans = 0;
memset(max, 0, sizeof max);
for(int i = n - 1; ~i; --i) {
int cur = 0;
for(int j = i + 1; j < n; ++j)
if(g[i][j]) adj[1][cur++] = j;
path[0] = i;
dfs(cur, 1);
max[i] = ans;
}
return ans;
}
} solver;
int n;
bool notPrime[N];
void gao() {
for(int i = 2; i * i < N; ++i) {
if(notPrime[i]) continue;
for(int j = i * i; j < N; j += i) notPrime[j] = true;
}
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
gao();
while(scanf("%d", &n) == 1) {
vector<int> a(n);
for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
solver.init(n);
for(int i = 0; i < n; ++i)
for(int j = i + 1; j < n; ++j)
solver.g[i][j] = solver.g[j][i] = !notPrime[a[i] + a[j]];
int ans = solver.maxClique();
printf("%d\n", ans);
for(int i = 0; i < ans; ++i)
printf("%d%c", a[solver.clique[i]], " \n"[i == ans - 1]);
}
return 0;
}


1. 除非注明,本博文即为原创,转载请注明链接地址
2. 本博文只代表博主当时的观点或结论,请不要恶意攻击
3. 如果本文帮到了您,不妨点一下 下面分享到 按钮,让更多的人看到