假设我们有一个整数排列 'seq' 和一个大小为 m 的整数对数组 'pairs'，其中包含整数 0 到 n - 1。现在，我们尽可能多地对 seq 执行以下操作，以使 seq[i] = i (0 ≤ i
我们必须选择一个整数 j，其中 0
我们必须找出 i 的最大值，使得在多次执行操作后 seq[i] = i。
因此，如果输入是 n = 4，m = 2，seq = {0, 3, 2, 1}，pairs = {{0, 1}, {2, 3}}，那么输出将是 2。
最大可能的值是 2。
为了解决这个问题，我们将按照以下步骤进行：
N := 100
Define an array tp of size: N.
Define arrays vtmp, vis of size N.
Define a function dfs(), this will take j, k,
tp[j] := k
insert j at the end of vtmp[k]
for each value b in vis[j], do:
if tp[b] is not equal to 0, then:
Ignore following part, skip to the next iteration
dfs(b, k)
res := 0
for initialize i := 0, when i < n, update (increase i by 1), do:
if seq[i] is same as i, then:
(increase res by 1)
for initialize i := 0, when i < m, update (increase i by 1), do:
a := first value of pairs[i]
b := second value of pairs[i]
insert b at the end of vis[a]
insert a at the end of vis[b]
idx := 1
for initialize i := 0, when i < n, update (increase i by 1), do:
if tp[i] is same as 0, then:
dfs(i, idx)
for each element j in vtmp[idx], do:
if tp[seq[j]] is same as idx and seq[j] is not equal to j, then:
(increase res by 1)
(increase idx by 1)
print(res)让我们看下面的实现以更好地理解−
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
#define N 100
int tp[N];
vector<int> vtmp[N], vis[N];
void dfs(int j, int k){
tp[j] = k;
vtmp[k].push_back(j);
for(auto b : vis[j]) {
if(tp[b] != 0)
continue;
dfs(b, k);
}
}
void solve(int n, int m, int seq[], vector<pair<int, int>> pairs) {
int res = 0;
for(int i = 0; i < n; i++){
if(seq[i] == i)
res++;
}
for(int i = 0; i < m; i++){
int a = pairs[i].first;
int b = pairs[i].second;
vis[a].push_back(b);
vis[b].push_back(a);
}
int idx = 1;
for(int i = 0; i < n; i++) {
if(tp[i] == 0) {
dfs(i, idx);
for(auto j: vtmp[idx]){
if(tp[seq[j]] == idx && seq[j] != j)
res++;
}
idx++;
}
}
cout<< res;
}
int main() {
int n = 4, m = 2, seq[] = {0, 3, 2, 1};
vector<pair<int,int>> pairs = {{0, 1}, {2, 3}};
solve(n, m, seq, pairs);
return 0;
}4, 2, {0, 3, 2, 1}, {{0, 1}, {2, 3}}2
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