ZOJ 1516 Uncle Tom's Inherited Land (二分图最大匹配)

解题思路：将相邻的两个节点分到二部图的两个集合中，然后跑最大匹配算法。

AC代码：

/*
@Author: wchhlbt
@Date: 2017/8/2
*/
#include <bits/stdc++.h>

#define Fori(x) for(int i=0;i<x;i++)
#define Forj(x) for(int j=0;j<x;j++)
#define maxn 100
#define inf 0x3f3f3f3f
#define ONES(x) __builtin_popcount(x)
#define _ << " " <<
using namespace std;

typedef long long ll ;
const double eps =1e-8;
const int mod = 1000000007;
typedef pair<int, int> P;
const double PI = acos(-1.0);
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};

inline int read(){ int num; scanf("%d",&num); return num;}

int a[105][105];

vector<P> e1,e2;
int n,m;

//注意给nx,ny赋值
int vis[maxn];//存储每个点是否被访问过
int g[maxn][maxn];//邻接矩阵
int cx[maxn];//cx[i]表示最终求得的最大匹配中与xi匹配的y顶点
int cy[maxn];
int nx,ny;//分别存储x集合和y集合中的顶点个数
int x[maxn];
int y[maxn];
bool f[maxn];//表示其属于哪个集合
int dfs(int u)
{
for(int v = 1; v<=ny; v++){//遍历Y集合
if(g[u][v] && !vis[v]){
vis[v] = 1;
if(cy[v]==-1 || dfs(cy[v])){
cx[u] = v;
cy[v] = u;
return 1;
}
}
}
return 0;
}

int MaxMatch()
{
int res = 0;
memset(cx,-1,sizeof cx);
memset(cy,-1,sizeof cy);
for(int i = 1; i<=nx; i++){//遍历X集合
if(cx[i]==-1){
memset(vis,0,sizeof vis);
res += dfs(i);
}
}
return res;
}

void init()
{
e1.clear();
e2.clear();
memset(a,0,sizeof a);
memset(g,0,sizeof g);
nx = ny = 0;
}

bool connect(int i, int j)
{
if(abs(e1[i].first - e2[j].first)==1 && abs(e1[i].second-e2[j].second)==0)
return true;
if(abs(e1[i].first - e2[j].first)==0 && abs(e1[i].second-e2[j].second)==1)
return true;
return false;
}

int main()
{
while(~scanf("%d%d",&n,&m) && n+m){
init();// e a g
int k = read();
for(int i = 0; i<k; i++){
int x = read(); int y = read();
a[x][y] = 1;
}
int cnt1 = 0, cnt2 = 0;
for(int i = 1; i<=n; i++){
for(int j = 1; j<=m; j++){
if(a[i][j]) continue;
if((i+j)%2==0)
e1.push_back(make_pair(i,j));
else
e2.push_back(make_pair(i,j));
}
}

for(int i = 0; i<e1.size(); i++){
for(int j = 0; j<e2.size(); j++){
if(connect(i,j)){
g[i+1][j+1] = 1;
}
}
}
nx = e1.size();
ny = e2.size();
int ans = MaxMatch();
cout << ans << endl;
}
return 0;
}