Codeforces 1186F - Vus the Cossack and a Graph 模拟乱搞/欧拉回路
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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\n", ans); for (int i = 0; i < m; i++) if(vis[i]) continue; printf("%d %d\n", 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\n", res); for (int i = 1; i <= totm; i++) if(vis[i] == 0) if(a[i].flag == 1) printf("%d %d\n", a[i].u, a[i].v); //6 6 //3 4 //4 5 //5 3 //1 3 //1 2 //2 3
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