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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; }
CodeForces 731C Socks (DFS或并查集)
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5971571.html