AW372 棋盘覆盖

题目地址

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易错点：

• 可以双向加边这个一定要赞一个.
• 在这种一次匹配则两个对象都被对方匹配上的题目中，应当在dfs成功时设置两个对象的match值为对方，即match[u]=v,match[v]=u.

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#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN=20000;
struct Edge{
	int from,to,nxt;
}e[MAXN*100];
int head[MAXN*100],edgeCnt=1;
void addEdge(int u,int v){
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
bool vis[MAXN];
int match[MAXN];
bool dfs(int x){
	for(int i=head[x];i;i=e[i].nxt){
		int nowV=e[i].to;
		if(vis[nowV])continue;
		vis[nowV]=1;
		if(!match[nowV]||dfs(match[nowV])){
			match[nowV]=x;
			match[x]=nowV;
			return 1;
		}
	}
	return 0;
}
int n;
int getHash(int i,int j){
	return (i-1)*n+j;
}
bool ban[110][110];
int main(){
	int t;
	scanf("%d%d",&n,&t);
	for(int i=1;i<=t;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		ban[x][y]=1;
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		if(ban[i][j])continue;
		int nowV=getHash(i,j);
		if(!ban[i-1][j]&&i-1>0)addEdge(nowV,getHash(i-1,j));
		if(!ban[i+1][j]&&i+1<=n)addEdge(nowV,getHash(i+1,j));
		if(!ban[i][j+1]&&j+1<=n)addEdge(nowV,getHash(i,j+1));
		if(!ban[i][j-1]&&j-1>0)addEdge(nowV,getHash(i,j-1));
	}
	int ans=0;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		int nowV=getHash(i,j);
		if(!match[nowV]){
			memset(vis,0,sizeof(vis));
			if(dfs(nowV))ans++;
		}
	}
	printf("%d
",ans);
	return 0;
}