CodeForces 731C Socks

Posted 2020-09-12 Lorazepam

http://codeforces.com/problemset/problem/731/C

并查集+贪心 将要求颜色相同的袜子序号放入一个集合中

贪心：然后统计这个集合中出现次数最多但颜色 可以得到这个集合要repain的次数

代码有难度

统计集合数

int tot;//总的集合数

for (int i = 1; i <= n; i++)

　　if(par[i] == i)

{

　　rec[tot++] = i;//记录根节点

} 

//统计每个集合红颜色的个数

map<int, int> clmp;

vector<int> G[MAXN];

for(int i = 1; i <= n; i++)

{

　　G[par[i]].push_back(color[i]);

}

int ans = 0;

for(int i = 0; i < tot; i++)//统计颜色情况

{

　　clmp.clear();

　　for (int j = 0; j < v[rec[i]].size(); j++)

　　clmp[v[rec[i]][j]]++;

}

 1 #include <bits/stdc++.h>
 2 #define MAXN 200007
 3 using namespace std;
 4 
 5 int par[MAXN];
 6 
 7 int find(int x)
 8 {
 9     if (par[x] == x) return x;
10     else return par[x] = find(par[x]);
11 }
12 bool same(int x, int y)
13 {
14     int px, py;
15     px = find(x);
16     py = find(y);
17     return px == py;
18 }
19 void unite(int x, int y)
20 {
21     int px = find(x);
22     int py = find(y);
23     par[px] = py;
24 }
25 int color[MAXN];
26 int rec[MAXN];
27 vector<int> v[MAXN];
28 map<int,int> clmap;
29 int main()
30 {
31     int day, paint, socks, ans = 0;
32     scanf("%d%d%d", &socks, &day, &paint);
33     for (int i =  0;  i <= socks; i++) par[i] = i;
34     for (int i = 1; i <= socks; i++)
35     {
36         scanf("%d", &color[i]);
37     }
38     for (int i = 0; i < day; i++)
39     {
40         int l, r;
41         scanf("%d%d", &l, &r);
42         unite(l, r);
43     }
44     int tot = 0;
45     for (int i = 1; i <= socks; i++)
46     {
47         if (i == find(i))
48         {
49             rec[tot++] = i;
50         }
51     }
52     for (int i = 1; i <= socks; i++)
53     {
54         v[par[i]].push_back(color[i]);
55     }
56     for (int i = 0; i < tot; i++)
57     {
58         clmap.clear();
59         for(int j = 0; j < v[rec[i]].size(); j++)
60         {
61             clmap[v[rec[i]][j]]++;
62         }
63         map<int,int> :: iterator it;
64         int maxn = 0;
65         for (it = clmap.begin(); it != clmap.end(); it++)
66         {
67             maxn = max(maxn, (*it).second);
68         }
69         ans += v[rec[i]].size() - maxn;
70     }
71     cout << ans << endl;
72 }