Uva 11419 - SAM I AM（最大匹配模板、找出最小覆盖点模板）

连接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=560&problem=2414&mosmsg=Submission+received+with+ID+11685942

 

Problem C
SAM I AM
Input: Standard Input

Output: Standard Output

 

The world is in great danger!! Mental's forces have returned to Earth to eradicate humankind. Our last hope to stop this great evil is Sam “Serious” Stone. Equipped with various powerful weapons, Serious Sam starts his mission to destroy the forces of evil.

After fighting two days and three nights, Sam is now in front of the temple KOPTOS where Mental's general Ugh Zan III is waiting for him. But this time, he has a serious problem. He is in shortage of ammo and a lot of enemies crawling inside the temple waiting for him. After rounding thetemple Sam finds that the temple is in rectangle shape and he has the locations of all enemies in the temple.

All of a sudden he realizes that he can kill the enemies without entering the temple using the great cannon ball which spits out a gigantic ball bigger than him killing anything it runs into and keeps on rolling until it finally explodes. But the cannonball can only shoot horizontally or vertically and all the enemies along the path of that cannon ball will be killed.

Now he wants to save as many cannon balls as possible for fighting with Mental. So, he wants to know the minimum number of cannon balls and the positions from which he can shoot the cannonballs to eliminate all enemies from outside that temple.

 

Input

Here, the temple is defined as a RXC grid. The first line of each test case contains 3 integers: R(0<R<1001), C(0<C<1001) representing the grid of temple (R means number of row and C means number of column of the grid) and the number of enemies N(0<N<1000001) inside the temple. After that there are N lines each of which contains 2 integers representing the position of the enemies in that temple. Each test case is followed by a new line (except the last one). Input is terminated when R=C=N=0. The size of the input file is around 1.3 MB.

 

Output

For each test case there will be one line output. First print the minimum number (m) of cannonballs needed to wipe out the enemies followed by a single space and then m positions from which he can shoot those cannonballs. For shooting horizontally print “r” followed by the row number and for vertical shooting print “c” followed by the column number. If there is more than one solution any one will do.

 

Sample Input                               Output for Sample Input

 

4 4 3 1 1 1 4 3 2   4 4 2 1 1 2 2   0 0 0

2 r1 r3 2 r1 r2

 

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 1000 + 5; // 单侧顶点的最大数目

// 二分图最大基数匹配
struct BPM {
  int n, m;               // 左右顶点个数
  vector<int> G[maxn];    // 邻接表
  int left[maxn];         // left[i]为右边第i个点的匹配点编号，-1表示不存在
  bool T[maxn];           // T[i]为右边第i个点是否已标记

  int right[maxn];        // 求最小覆盖用
  bool S[maxn];           // 求最小覆盖用

  void init(int n, int m) {
    this->n = n;
    this->m = m;
    for(int i = 0; i < n; i++) G[i].clear();
  }

  void AddEdge(int u, int v) {
    G[u].push_back(v);
  }

  bool match(int u){
    S[u] = true;
    for(int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (!T[v]){
        T[v] = true;
        if (left[v] == -1 || match(left[v])){
          left[v] = u;
          right[u] = v;
          return true;
        }
      }
    }
    return false;
  }

  // 求最大匹配
  int solve() {
    memset(left, -1, sizeof(left));
    memset(right, -1, sizeof(right));
    int ans = 0;
    for(int u = 0; u < n; u++) { // 从左边结点u开始增广
      memset(S, 0, sizeof(S));
      memset(T, 0, sizeof(T));
      if(match(u)) ans++;
    }
    return ans;
  }

  // 求最小覆盖。X和Y为最小覆盖中的点集
  int mincover(vector<int>& X, vector<int>& Y) {
    int ans = solve();
    memset(S, 0, sizeof(S));
    memset(T, 0, sizeof(T));
    for(int u = 0; u < n; u++)
      if(right[u] == -1) match(u); // 从所有X未盖点出发增广
    for(int u = 0; u < n; u++)
      if(!S[u]) X.push_back(u); // X中的未标记点
    for(int v = 0; v < m; v++)
      if(T[v]) Y.push_back(v);  // Y中的已标记点
   return ans;
  }
};

BPM solver;

int R, C, N;

int main(){
  int kase = 0;
  while(scanf("%d%d%d", &R, &C, &N) == 3 && R && C && N) {
    solver.init(R, C);
    for(int i = 0; i < N; i++) {
      int r, c;
      scanf("%d%d", &r, &c); r--; c--;
      solver.AddEdge(r, c);
    }
    vector<int> X, Y;
    int ans = solver.mincover(X, Y);//把结果放在X,Y里面
    printf("%d", ans);
    for(int i = 0; i < X.size(); i++) printf(" r%d", X[i]+1);
    for(int i = 0; i < Y.size(); i++) printf(" c%d", Y[i]+1);
    printf("\n");
  }
  return 0;
}