【圖論02】二分圖 1001 過山車

演算法思路：二分圖。

簡單的二分圖求最大匹配的模板題，匈牙利演算法。任意選擇一個點做為起點開始遍曆，如果找到“增廣路”，則sum++（sum初始化為0）。對所有的點嘗試過一次找“增廣路”之後（共做了N次），輸出sum即為所求。

 

//模板開始#include <string>   #include <vector>   #include <algorithm>   #include <iostream>   #include <sstream>   #include <fstream>   #include <map>   #include <set>   #include <cstdio>   #include <cmath>   #include <cstdlib>   #include <ctime>#include <iomanip>#include <string.h>#include <queue>#define SZ(x) (int(x.size()))using namespace std;int toInt(string s){istringstream sin(s); int t; sin>>t; return t;}template<class T> string toString(T x){ostringstream sout; sout<<x; return sout.str();}typedef long long int64;int64 toInt64(string s){istringstream sin(s); int64 t; sin>>t;return t;}template<class T> T gcd(T a, T b){ if(a<0) return gcd(-a, b);if(b<0) return gcd(a, -b);return (b == 0)? a : gcd(b, a % b);}//模板結束（通用部分）#define ifs cin#define MAX 1010int juzhen[MAX][MAX];int used[MAX];int mat[MAX];void init(){memset(juzhen, 0, sizeof(juzhen));}int Augment(int s, int n, int x){int i;for(i = s; i <= n; i++){if(!used[i] && juzhen[x][i]){used[i] = 1;if(mat[i] == -1 || Augment(s, n, mat[i])){mat[i] = x;return 1;}}}return 0;}int Hungary(int s, int n){int i, sum = 0;memset(mat, -1, sizeof(mat));for(i = s; i <= n; i++){memset(used, 0, sizeof(used));if(Augment(s, n, i)){sum++;}}return sum;}//【圖論02】二分圖 1001 過山車int main(){//ifstream ifs("shuju.txt", ios::in);int k, m, n;int a, b;while(ifs>>k && k != 0){ifs>>m>>n;memset(juzhen, 0, sizeof(juzhen));for(int i = 0; i < k; i++){ifs>>a>>b;juzhen[a][m + b] = 1;//juzhen[m + b][a] = 1;}int max_match = Hungary(1, m + n);cout<<max_match<<endl;}return 0;}