poj1469-COURSES

poj1469-COURSES

COURSES
Time Limit: 1000MS
Memory Limit: 10000K
Description
Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions:
every student in the committee represents a different course (a student can represent a course if he/she visits that course)
each course has a representative in the committee
Input
Your program should read sets of data from the std input. The first line of the input contains the number of the data sets. Each data set is presented in the following format:
P N
Count1 Student1 1 Student1 2 … Student1 Count1
Count2 Student2 1 Student2 2 … Student2 Count2
…
CountP StudentP 1 StudentP 2 … StudentP CountP
The first line in each data set contains two positive integers separated by one blank: P (1 <= P <= 100) - the number of courses and N (1 <= N <= 300) - the number of students. The next P lines describe in sequence of the courses �from course 1 to course P, each line describing a course. The description of course i is a line that starts with an integer Count i (0 <= Count i <= N) representing the number of students visiting course i. Next, after a blank, you抣l find the Count i students, visiting the course, each two consecutive separated by one blank. Students are numbered with the positive integers from 1 to N.
There are no blank lines between consecutive sets of data. Input data are correct.
Output
The result of the program is on the standard output. For each input data set the program prints on a single line “YES” if it is possible to form a committee and “NO” otherwise. There should not be any leading blanks at the start of the line.
Sample Input
2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1
Sample Output
YES
NO
Source
Southeastern Europe 2000

二分圖匹配
解法1：匈牙利算法 O（Esqrt（v））500MS

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
const int MAXN = 400;

int read(){
    int k = 1, x = 0; char ch = getchar();
    while (ch<'0'||ch>'9') { if(ch == '-') k = -1; ch = getchar(); }
    while ('0'<=ch&&ch<='9') { x=x*10+ch-'0'; ch = getchar();}
    return k*x;
}

vector<int> g[MAXN];
int used[MAXN], from[MAXN];
int p, n;

bool match(int u) {//cout<<u<<endl;
    for (int i = 0; i < g[u].size(); ++i) {
      if (!used[g[u][i]]){
        used[g[u][i]] = 1;
        if (from[g[u][i]] == -1 || match(from[g[u][i]])) {
            from[g[u][i]] = u;
          return 1;
        }
      }
    }
    return 0;
}

int hungary(){
    int res = 0;
    memset(from, 255, sizeof from);
    for (int i = 1; i <= n; ++i) {
      memset(used, 0, sizeof used);
      if (match(i)) 
        res++;
    }
    return res;
}

int main()
{
    int T; T = read();
    while (T--) {
      p = read(), n = read();
      for (int i = 1; i <= max(p, n); ++i) g[i].clear();
      for (int i = 1; i <= p; ++i) {
        int count = read(), x;
        for (int j = 1; j <= count; ++j){
          x = read();
          g[i].push_back(x);
        }
      }
      if (hungary() == p) cout<<"YES\n";
      else cout<<"NO\n";
    }
    return 0;
} 

解法2：Hopcroft-Karp算法 O（Esqrt（v））485MS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 400;

int read(){
    int k = 1, x = 0; char ch = getchar();
    while (ch<'0'||ch>'9') {
      if (ch == '-') k = -1;
      ch = getchar();
    }
    while ('0'<=ch && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return k*x;
}

int dx[MAXN], dy[MAXN], mx[MAXN], my[MAXN], vis[MAXN], n, p;
vector<int> g[MAXN];
queue<int> que;

bool find(int u){//cout<<u<<endl;
    for (int i = 0; i < g[u].size(); ++i)
      if (!vis[g[u][i]] && dy[g[u][i]] == dx[u] + 1) {
        vis[g[u][i]] = 1;
        if (!my[g[u][i]] || find(my[g[u][i]])) {
          mx[u] = g[u][i];
          my[g[u][i]] = u;
          return 1;
        }
      }
    return 0;
}

int matching(){
    memset(mx, 0, sizeof mx);
    memset(my, 0, sizeof my);
    int res = 0;
    while (1) {
      bool flag = false;
      memset(dx, 0, sizeof dx);
      memset(dy, 0, sizeof dy);
      while (!que.empty()) que.pop();
      for (int i = 1; i <= p; ++i)
        if (!mx[i])
          que.push(i);
      while (!que.empty()) {
        int u = que.front();
        que.pop();
        for (int i = 0; i < g[u].size(); ++i)
          if (!dy[g[u][i]]) {
            dy[g[u][i]] = dx[u] + 1;
            if (my[g[u][i]]) {
              dx[my[g[u][i]]] = dy[g[u][i]] + 1;
              que.push(my[g[u][i]]);
            } else
              flag = true;
          }
      }
      if (!flag) break;
      memset(vis, 0, sizeof vis);
      for (int i = 1; i <= p; ++i) 
        if (!mx[i] && find(i))
          res++;
    }
    return res;
}

int main()
{
    int T; T = read();
    while (T--) {
      p = read(), n = read();
      for (int i = 1; i <= p; ++i) g[i].clear();
      for (int i = 1; i <= p; ++i) {
        int count = read(), x;
        for (int j = 1; j <= count; ++j){
          x = read();
          g[i].push_back(x);
        }
      }
      if (matching() == p) cout<<"YES\n";
      else cout<<"NO\n";
    }
    return 0;
}