Codeforces 1186F

题意：给你一张无向图，要求对这张图进行删边操作，要求删边之后的图的总边数 >= ceil((n + m) / 2), 每个点的度数 >= ceil(deg[i] / 2)。（deg[i]是原图中i的度数）

思路1：模拟 + 乱搞

直接暴力删就行了，读入边之后随机打乱一下就很难被hack了。

代码：

#include <bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define db double
#define pii pair<int, int>
using namespace std;
const int maxn = 1000010;
struct node {
	int u, v, id;
};
vector<node> G;
int deg[maxn], limit[maxn], a[maxn];
pii b[maxn];
const LL mod = 1e9 + 7;
LL add(LL x, LL y) {return (x + y) % mod;}
LL mul(LL x, LL y) {return (x * y) % mod;}
bool vis[maxn];
bool cmp(int x, int y) {
	return deg[x] > deg[y];
}
int main() {
	int n, m, u, v;
	scanf("%d%d", &n, &m);
	int tot_limit = (n + m + 1) / 2;
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &u, &v);
//		G[u].push_back((node){u, v, i});
//		G[v].push_back((node){v, u, i});
		G.push_back((node){u, v, i});
		deg[u]++;
		deg[v]++;
	}
	for (int i = 1; i <= n; i++) {
		limit[i] = (deg[i] + 1) / 2;
	}
	random_shuffle(G.begin(), G.end());
	int ans = m;
	for (int j = 0; j < G.size() && ans > tot_limit; j++) {
		int v = G[j].v, now = G[j].u;
		if(deg[v] == limit[v]) continue;
		if(deg[now] == limit[now]) continue;
		vis[j] = 1;
		ans--;
		deg[v]--;
		deg[now]--;
	}
	printf("%d
", ans);
	for (int i = 0; i < m; i++) {
		if(vis[i]) continue;
		printf("%d %d
", G[i].u, G[i].v);
	}
}

思路2（官方题解）：新建0号点，把0号点和图中所有度数为奇数的点相连，形成一张新图。在新图上跑一遍欧拉回路，把欧拉回路记录的边中偶数位置的删掉，删的时候如果是新加的边，就直接删了。否则，看一下这条边相邻的两条边是不是新加的边并且可以删，如果可以，那就删新加的边，否则删这条边。即迫不得已的情况才会删除原图的边。

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
struct edge {
	int u, v, flag;
};

int st[maxn * 2], ans[maxn * 2], re[maxn * 2];
edge a[maxn * 2];
int head[maxn], id[maxn * 4], Next[maxn * 4], ver[maxn * 4], tot, totm, tot_ans;
bool v[maxn * 4], vis[maxn * 4];
int deg[maxn];
int Top;
void add(int x, int y, int z) {
	ver[++tot] = y, id[tot] = z, Next[tot] = head[x], head[x] = tot;
}
void euler (int s) {
	tot_ans = 0;
	st[++Top] = s;
	while(Top > 0) {
		int x = st[Top], i = head[x];
		while(i && v[i]) i = Next[i];
		if(i) {
			st[++Top] = ver[i];
			re[Top] = id[i];
			v[i] = v[i ^ 1] = 1;
			head[x] = Next[i];
		} else {
			ans[++tot_ans] = re[Top];
			Top--;
		}
	}
}
int main() {
	int n, m, x, y;
	scanf("%d%d", &n, &m);
	tot = 1;
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &x, &y);
		totm++;
		a[totm] = (edge){x, y, 1}; 
		add(x, y, totm), add(y, x, totm);
		deg[x]++, deg[y]++;
	}
	for (int i = 1; i <= n; i++) {
		if(deg[i] & 1) {
			totm++; 
			a[totm] = (edge){0, i, 0}; 
			add(0, i, totm), add(i, 0, totm);
		}
	}
	int res = m;
	for (int i = 0; i <= n; i++) {
		euler(i);
		for (int j = 2; j <= tot_ans; j += 2) {
			int now = ans[j];
			if(a[now].flag == 0) vis[now] = 1;
			else {
				int tmp = ans[j - 1];
				if(a[tmp].flag == 0 && vis[tmp] == 0) {
					vis[tmp] = 1;
					continue;
				}
				int Next = j + 1;
				if(j == tot_ans) Next = 1;
				tmp = ans[Next];
				if(a[tmp].flag == 0 && vis[tmp] == 0) {
					vis[tmp] = 1;
					continue;
				}
				vis[now] = 1;
				res--;
			}
		}
	}
	printf("%d
", res);
	for (int i = 1; i <= totm; i++) {
		if(vis[i] == 0) {
			if(a[i].flag == 1) {
				printf("%d %d
", a[i].u, a[i].v);
			}
		}
	}
}
//6 6
//3 4
//4 5
//5 3
//1 3
//1 2
//2 3