CodeForces 731C Socks (DFS或并查集)

题意：有n只袜子，k种颜色，在m天中，问最少修改几只袜子的颜色，可以使每天穿的袜子左右两只都同颜色。

析：很明显，每个连通块都必须是同一种颜色，然后再统计最多颜色的就好了，即可以用并查集也可以用DFS。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
bool vis[maxn];
map<int, int> mp;
vector<int> G[maxn];
int cnt, mmax;

void dfs(int u){
    mmax = Max(mmax, mp[a[u]]);
    ++cnt;
    for(int i = 0; i < G[u].size(); ++i){
        int v = G[u][i];
        if(vis[v]) continue;
        ++mp[a[v]];
        vis[v] = true;
        dfs(v);
    }
}

int main(){
    int k;
    while(scanf("%d %d %d", &n, &m, &k) == 3){
        for(int i = 1; i <= n; ++i)  scanf("%d", a+i), G[i].clear();
        int u, v;
        for(int i = 0; i < m; ++i){
            scanf("%d %d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        int ans = 0;
        memset(vis, false, sizeof vis);
        for(int i = 1; i <= n; ++i) if(!vis[i] && G[i].size()){
            mp.clear();  cnt = mmax = 0;
            vis[i] = true;
            ++mp[a[i]];
            dfs(i);
            ans += cnt - mmax;
        }
        printf("%d\n", ans);
    }
    return 0;
}