POJ 3710 无向图简单环树上删边

Posted 2020-10-12 ( m Lweleth)

结论题，这题关键在于如何转换环，可以用tarjan求出连通分量后再进行标记，也可以DFS直接找到环后把点的SG值变掉就行了

/** @Date    : 2017-10-23 19:47:47
  * @FileName: POJ 3710 简单环 树上删边 DFS.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
#include <math.h>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e3 + 20;
const double eps = 1e-8;

struct node{
	int nxt, to;
}edg[N*2];

int vis[N];
int head[N*2];
int tot;
int st[10010], top;

int sg[N];

void init()
{
	MMG(head);
	MMG(vis);	
	MMF(sg);
	tot = top = 0;
}

inline void add(int x, int y)
{
	edg[tot].nxt = head[x];
	edg[tot].to = y;
	head[x] = tot++;
}

void dfs(int x, int pre)
{
	st[++top] = x;
	vis[x] = 1;
	int flag = 0;
	for(int i = head[x]; ~i; i = edg[i].nxt)
	{
		if(edg[i].to == pre && !flag)
		{
			flag = 1;
			continue;
		}
		if(vis[edg[i].to] == 1)
		{
			int nw = st[top];
			int cnt = 1;
			while(nw != edg[i].to)
				vis[nw] = 0, nw = st[--top], cnt++;
			if(cnt & 1)//奇数环变边
				sg[edg[i].to] ^= 1;
		}
		else if(vis[edg[i].to] == -1)
		{
			dfs(edg[i].to, x);
			if(vis[edg[i].to])
				sg[x] ^= sg[edg[i].to] + 1;
		}
	}
	if(vis[x])
		top--;
}
int main()
{
	int T;
	while(cin >> T){//多组样例233
		int ans = 0;
		while(T--)
		{
			int n, m;
			scanf("%d%d", &n, &m);
			init();
			for(int i = 1; i <= m; i++)
			{
				int x, y;
				scanf("%d%d", &x, &y);
				add(x, y);
				add(y, x);
			}
			dfs(1, -1);
			ans ^= sg[1];
		}
		printf("%s\n", ans?"Sally":"Harry");
	}
    return 0;
}