Bridges
题目背景：
bzoj2095
分析：今天get到一种新技能，混合图欧拉回路，首先我们来考虑下普通的有向图，当且仅当每一个点的入度等于出度时，存在一个欧拉回路，无向图则是每一个点的度数为偶数，现在我们考虑混合图，首先对于有向边直接选择，对于无向边，我们随便定向，之后如果原图存在欧拉回路，那么每一个点的入度和出度之差一定是偶数（若有一条边方向相反，则会有两个点的入度和出度之差的绝对值加2），考虑调整的方式，我们选择网络流，对于每一个入度大于出度的点，从源点向当前点连接边权为（入度 – 出度）/ 2的边，对于每一个出度大于入度的点，从当前点向汇点连边权为（出度 – 入度）/ 2的边，对于每一条我们定过向的无向边，反向连一条边权为1的边，表示经过一次这条边就表示将这条有向边反向，调整2的入度和出度。然后直接二分能够承受的风力大小，建图跑最大流，判断获得的最大流是否为之前连向起点的权值和，若满足则存在否则不存在。注意原图可能不连通，所以我还搞了个并查集，事实上应该不用······
Source：
/*created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;
}///*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = read(), iosig = false; !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true; }for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;
}template<class T>
inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}
}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);
}/*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = getchar(), iosig = false; !isdigit(c); c = getchar())if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int MAXN = 1000 + 10;
const int INF = ~0u >> 1;struct node {int to, w, f, rev;node(int to = 0, int w = 0, int rev = 0) : to(to), w(w), f(0), rev(rev) {}
} ;struct edge {int a, b, c, d;edge(int a = 0, int b = 0, int c = 0, int d = 0) : a(a), b(b), c(c), d(d) {}
} edges[MAXN << 1];std::vector<node> edge[MAXN];int n, m, a, b, c, d, s, t;
int dis[MAXN], temp[MAXN], gap[MAXN], father[MAXN];
int out_degree[MAXN], in_degree[MAXN];inline void add_edge(int x, int y, int w) {edge[x].push_back(node(y, w, edge[y].size()));edge[y].push_back(node(x, 0, edge[x].size() - 1));
}inline void read_in() {R(n), R(m), s = 0, t = n + 1;for (int i = 1; i <= m; ++i)R(edges[i].a), R(edges[i].b), R(edges[i].c), R(edges[i].d);
}inline int sap(int cur, int flow, int s, int t, int n) {if (cur == t) return flow;int del = 0;static int temp[MAXN];for (int p = temp[cur]; p < edge[cur].size(); ++p) {node *e = &edge[cur][p];if (e->w > 0 && dis[e->to] + 1 == dis[cur]) {int ret = sap(e->to, std::min(e->w, flow - del), s, t, n);e->w -= ret, edge[e->to][e->rev].w += ret, temp[cur] = p;if ((del += ret) == flow || dis[cur] >= n) return temp[cur] = 0, del;}}if (--gap[dis[cur]] == 0) dis[s] = n;gap[++dis[cur]]++, temp[cur] = 0;return del;
}inline int sap(int s, int t, int n) {int ret = 0;memset(gap, 0, sizeof(int) * (n + 1));memset(dis, 0, sizeof(int) * (n + 1));for (gap[0] = n; dis[s] < n; ) ret += sap(s, ~0u >> 1, s, t, n);return ret;
}inline int get_father(int x) {return (father[x] == x) ? x : (father[x] = get_father(father[x]));
}inline void unite(int x, int y) {int fa1 = get_father(x), fa2 = get_father(y);if (fa1 != fa2) father[fa1] = fa2;
}inline bool check(int mid) {for (int i = 0; i <= n + 1; ++i) edge[i].clear(), father[i] = i, in_degree[i] = 0, out_degree[i] = 0;for (int i = 1; i <= m; ++i) {int u = edges[i].a, v = edges[i].b;if (edges[i].c <= mid && edges[i].d <= mid)out_degree[u]++, in_degree[v]++, add_edge(v, u, 1), unite(u, v);else if (edges[i].c <= mid) out_degree[u]++, in_degree[v]++, unite(u, v);else if (edges[i].d <= mid) out_degree[v]++, in_degree[u]++, unite(u, v);}int ans = 0, fa = get_father(1), tt = 0;for (int i = 1; i <= n; ++i) {if (get_father(i) != fa) return false;int x = in_degree[i] - out_degree[i];if (abs(x) & 1) return false;if (x > 0) add_edge(s, i, x / 2), ans += x / 2;else if (x < 0) add_edge(i, t, -x / 2), tt -= x / 2; } return (sap(s, t, t + 1) == ans);
}inline void binary() {int l = -1, r = 1000 + 1;while (l + 1 < r) {int mid = l + r >> 1;check(mid) ? (r = mid) : (l = mid);}if (r == 1000 + 1) std::cout << "NIE";else std::cout << r;
}int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);read_in();binary();return 0;
}