UVA 11865 Stream My Contest (二分+最小树形图)

题意：给定一个网络，一个服务器，其他的是客户机，有 m 条连线，每条有一个带宽和花费（单向边），让你用不超过 c 的花费，使得 0 到 所有的机器都能到达，并且使得最小带宽最大。

析：很明显是二分题，然后在判断，就是保证从 0 到所有的点都是通路，这就是最小树形图，直接上模板就好。

代码如下;

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 60 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
  int u, v, cost;
};
Edge edge[10050], a[10050];
int b[10050];
int pre[maxn], id[maxn], vis[maxn], in[maxn];

int solve(int rt, int n, int m){
  int ans = 0;
  while(true){
    ms(in, INF);
    FOR(i, 0, m)  if(edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v]){
      pre[edge[i].v] = edge[i].u;
      in[edge[i].v] = edge[i].cost;
    }
    FOR(i, 0, n)  if(i != rt && in[i] == INF)  return -1;
    int tn = 0;
    ms(id, -1);  ms(vis, -1);
    in[rt] = 0;
    for(int i = 0; i < n; ++i){
      ans += in[i];
      int v = i;
      while(vis[v] != i && id[v] == -1 && v != rt){
        vis[v] = i;
        v = pre[v];
      }
      if(v != rt && id[v] == -1){
        for(int u = pre[v]; u != v; u = pre[u])  id[u] = tn;
        id[v] = tn++;
      }
    }
    if(tn == 0)  break;
    FOR(i, 0, n)  if(id[i] == -1)  id[i] = tn++;
    for(int i = 0; i < m; ){
      int v = edge[i].v;
      edge[i].u = id[edge[i].u];
      edge[i].v = id[edge[i].v];
      if(edge[i].u != edge[i].v)  edge[i++].cost -= in[v];
      else swap(edge[i], edge[--m]);
    }
    n = tn;
    rt = id[rt];
  }
  return ans;
}

bool judge(int mid, int c){
  int cnt = 0;
  for(int i = 0; i < m; ++i)
    if(b[i] >= mid)  edge[cnt++] = a[i];
  int ans = solve(0, n, cnt);
  if(ans == -1)  return false;
  bool ok = ans <= c;
  return ans <= c;
}

int main(){
  int T;  cin >> T;
  while(T--){
    int c;
    scanf("%d %d %d", &n, &m, &c);
    int l = 1, r = 0;
    for(int i = 0; i < m; ++i)  scanf("%d %d %d %d", &a[i].u, &a[i].v, &b[i], &a[i].cost), r = max(r, b[i]);
    if(!judge(0, c)){ puts("streaming not possible.");  continue; }
    while(l <= r){
      int m = l + r >> 1;
      if(judge(m, c))  l = m + 1;
      else r = m - 1;
    }
    printf("%d kbps
", l - 1);
  }
  return 0;
}