hiho1122_二分图匈牙利算法

题目

    给定一个图的N个节点和节点之间的M条边，数据保证该图可以构成一个二分图。求该二分图最大匹配。 
题目链接：二分图最大匹配 
    首先通过染色法，将图的N个节点分成两个部分；然后通过匈牙利算法求二分图的最大匹配。

实现

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<algorithm>
using namespace std;

struct Edge{
	int to;
	int next;
};
Edge gEdges[80005];
int gEdgeIndex;
int gHead[10005];
int gMatch[10005];
bool gVisited[10005];
int gColor[10005];
bool conflict;
void Init(){
	gEdgeIndex = 0;
	memset(gHead, -1, sizeof(gHead));
	memset(gVisited, false, sizeof(gVisited));
	memset(gMatch, -1, sizeof(gMatch));
	memset(gColor, -1, sizeof(gColor));
	conflict = false;
}
void InsertEdge(int u, int v){
	int e = gEdgeIndex++;
	gEdges[e].to = v;
	gEdges[e].next = gHead[u];
	gHead[u] = e;
}

void PaintColor(int node, int c){
	if (conflict)
		return;
	gColor[node] = c;
	for (int e = gHead[node]; e != -1; e = gEdges[e].next){
		int x = gEdges[e].to;
		if (gColor[x] == c){
			conflict = true;
			return;
		}
		if (gColor[x] == -1)
			PaintColor(x, 1 - c);
	}
}

bool Dfs(int node){		
	for (int e = gHead[node]; e != -1; e = gEdges[e].next){
		int v = gEdges[e].to;
		if (!gVisited[v]){  
			gVisited[v] = true; //置位
			if (gMatch[v] == -1){
			//如果对方的阵营中v点还没有匹配，则成功找到一条交错路径
				gMatch[node] = v;
				gMatch[v] = node;
				return true;
			}
			else if (Dfs(gMatch[v])){
			//如果对方阵营中v有匹配点，匹配点肯定在己方阵营，则继续从己方阵营的点
			//递归
				gMatch[node] = v;
				gMatch[v] = node;
				return true;
			}
		}
	}
	return false;
}
int main(){
	int n, m, u, v;
	Init();
	scanf("%d %d", &n, &m);

	for (int i = 0; i < m; i++){
		scanf("%d %d", &u, &v);
		InsertEdge(u, v);
		InsertEdge(v, u);
	}
		
	for (int i = 1; i <= n; i++)
		if (gColor[i] == -1)
			PaintColor(i, 0);

	int ans = 0;
	for (int i = 1; i <= n; i++){
		if (gColor[i] == 0 && gMatch[i] == -1){//从固定一方阵营中，不断找未匹配点
		//寻找交错路，找到一条交错路，则匹配个数增加1
			memset(gVisited, false, sizeof(gVisited));
			if (Dfs(i))
				ans++;
		}
	}
	printf("%d
", ans);
	return 0;
}