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。
#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\n", ans);
return 0;
}
原文:http://www.cnblogs.com/gtarcoder/p/5612699.html