bzoj2095 Bridges

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;}