【BZOJ2095】【最大流】[Poi2010]Bridges 题解

Description

YYD为了减肥，他来到了瘦海，这是一个巨大的海，海中有n个小岛，小岛之间有m座桥连接，两个小岛之间不会有两座桥，并且从一个小岛可以到另外任意一个小岛。现在YYD想骑单车从小岛1出发，骑过每一座桥，到达每一个小岛，然后回到小岛1。霸中同学为了让YYD减肥成功，召唤了大风，由于是海上，风变得十分大，经过每一座桥都有不可避免的风阻碍YYD，YYD十分ddt，于是用泡芙贿赂了你，希望你能帮他找出一条承受的最大风力最小的路线。

Input

输入：第一行为两个用空格隔开的整数n（2<=n<=1000），m(1<=m<=2000)，接下来读入m行由空格隔开的4个整数a，b（1<=a,b<=n,a<>b），c，d(1<=c,d<=1000)，表示第i+1行第i座桥连接小岛a和b，从a到b承受的风力为c，从b到a承受的风力为d。

Output

输出：如果无法完成减肥计划，则输出NIE，否则第一行输出承受风力的最大值（要使它最小)

Sample Input

4 4

1 2 2 4

2 3 3 4

3 4 4 4

4 1 5 4

Sample Output

4

HINT

注意：通过桥为欧拉回路

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <set>#include <queue>#include <algorithm>#include <vector>#include <cstdlib>#include <cmath>#include <ctime>#include <stack>#define INF 2147483647#define LL long long#define clr(x) memset(x, 0, sizeof x)#define ms(a, x) memset(x, a, sizeof x)#define digit (ch <  '0' || ch >  '9')#define maxx(a, b, c) max(a, max(b, c))using namespace std;template <class T> inline void read(T &x) {    int flag = 1; x = 0;    register char ch = getchar();    while( digit) { if(ch == '-')  flag = -1; ch = getchar(); }    while(!digit) { x = (x<<1)+(x<<3)+ch-'0'; ch = getchar(); }    x *= flag;}const int maxm = 2005;int n,m;int a[maxm],b[maxm],c[maxm],d[maxm];int fa[maxm],size[maxm],cnt,maxn;struct edge { int to,f,nxt; } e[1001001];int head[maxm],dpt[maxm],tot = 1;int find(int x) {    if(!fa[x]) fa[x] = x, size[x] = 1;    return fa[x] == x ? x : fa[x] = find(fa[x]);}inline void unionn(int x, int y) {    x = find(x); y = find(y);      if(x == y) return ;    if(size[x] > size[y]) swap(x, y);    fa[x] = y; size[y] += size[x]; cnt--;}inline void add(int x, int y, int z) { e[++tot].to = y; e[tot].f = z; e[tot].nxt = head[x]; head[x] = tot; }bool bfs() {    int q[maxm], r = 0, h = 0;    ms(-1, dpt); q[++r] = 0; dpt[0] = 1;    while(r != h) {        int x = q[++h];          for(int i = head[x]; i; i = e[i].nxt) if(e[i].f && !~dpt[e[i].to]) {            dpt[e[i].to] = dpt[x]+1, q[++r] = e[i].to;            if(e[i].to == maxm-1) return true;        }    }    return false;}int dinic(int x, int flow) {      int left = flow;      if(x == maxm-1) return flow;    for(int tmp, i = head[x]; i && left; i = e[i].nxt) if(e[i].f && dpt[e[i].to] == dpt[x]+1)        tmp = dinic(e[i].to, min(left, e[i].f)), left -= tmp, e[i].f -= tmp, e[i^1].f += tmp;    if(left) dpt[x] = -1;    return flow-left;}inline bool check(int x) {    int degree[maxm]; cnt = n; tot = 1;    clr(head); clr(fa); clr(degree);    for(int i = 1; i <= m; i++)        if(d[i] <= x) unionn(a[i], b[i]), degree[a[i]]++, degree[b[i]]--, add(a[i], b[i], 1), add(b[i], a[i], 0);        else if(c[i] <= x) unionn(a[i], b[i]), degree[a[i]]--, degree[b[i]]++;    if(cnt >= 2) return false;    for(int i = 1; i <= n; i++) {        if(degree[i]&1) return false;        if(degree[i] > 0) add(0, i, degree[i]>>1), add(i, 0, 0);        else add(i, maxm-1, -degree[i]>>1), add(maxm-1, i, 0);    }    while(bfs()) dinic(0, INF);      for(int i = head[0]; i; i = e[i].nxt) if(e[i].f) return false;    return true;}int bisection(int l, int r) {    while(l+1 < r) {        int mid = l+r>>1;        if(check(mid)) r = mid;        else l = mid;    }    return check(l) ? l : r;}int main() {    read(n); read(m);    for(register int i = 1; i <= m; i++) {        read(a[0]); read(b[0]); read(c[0]); read(d[0]);        if(c[0] > d[0]) swap(a[0], b[0]), swap(c[0], d[0]);        a[i] = a[0]; b[i] = b[0];        c[i] = c[0]; d[i] = d[0];        maxn = maxx(maxn, c[0], d[0]);    }    int ans = bisection(1, maxn+1);    printf(ans == maxn+1 ? "NIE" : "%d",ans);    return 0;}