CodeForces 755C PolandBall and Forest (并查集)

题意:给定n个数,Ai的下标为1~n。对于每一个i,Ai与i在同一个树上,且是与i最远的点中id最小的点(这个条件变相的说明i与Ai连通)。求森林中树的个数。

分析:若i与Ai连通,则在同一个树上,因此连通块的个数就是树的个数。并查集即可。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int fa[MAXN];
int a[MAXN];
set<int> s;
int Find(int v){
    return fa[v] = (fa[v] == v) ? v : Find(fa[v]);
}
int main(){
    for(int i = 0; i < MAXN; ++i){
        fa[i] = i;
    }
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i){
        scanf("%d", &a[i]);
        int tx = Find(i);
        int ty = Find(a[i]);
        if(tx < ty) fa[ty] = tx;
        else fa[tx] = ty;
    }
    for(int i = 1; i <= n; ++i){
        if(Find(i) == i) s.insert(i);
        if(Find(a[i]) == a[i]) s.insert(a[i]);
    }
    printf("%d\n", s.size());
    return 0;
}
原文地址:https://www.cnblogs.com/tyty-Somnuspoppy/p/6294080.html