CF1037E Trips

题目链接：传送门

很好的思路
读完题目想一想
倒着建边会比较好处理
因为删边的处理比加边要容易
所以离线下来倒着搞就可以了

/*** @Date:   2019-04-14T10:03:39+08:00* @Last modified time: 2019-04-14T10:03:39+08:00*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define A 1000010
#define B 2010using namespace std;
typedef long long ll;
struct node {int next, to;
}e[A];
int head[A], num;
void add(int fr, int to) {e[++num].next = head[fr];e[num].to = to;head[fr] = num;
}
int ans, n, m, k, del[A], in[A], a[A], b[A], out[A];
map<pair<int, int>, bool> mm;
void work(int x) {queue<int> q; q.push(x);if (in[x] >= k or del[x]) return;del[x] = 1; ans--;while (!q.empty()) { //踢出能遍历到的所有不符合条件的点int fr = q.front(); q.pop();for (int i = head[fr]; i; i = e[i].next) {int ca = e[i].to;if (del[ca]) continue;if (!mm[make_pair(fr, ca)]) in[ca]--;if (in[ca] < k) {ans--; q.push(ca); del[ca] = 1;}}}
}int main(int argc, char const *argv[]) {cin >> n >> m >> k; ans = n;for (int i = 1; i <= m; i++) {cin >> a[i] >> b[i];add(a[i], b[i]); add(b[i], a[i]);in[a[i]]++; in[b[i]]++;}for (int i = 1; i <= n; i++) work(i);for (int i = m; i >= 1; i--) {out[i] = ans;if (!del[a[i]]) in[b[i]]--;if (!del[b[i]]) in[a[i]]--;mm[make_pair(a[i], b[i])] = mm[make_pair(b[i], a[i])] = 1;work(a[i]); work(b[i]);}for (int i = 1; i <= m; i++) cout << out[i] << endl;
}

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