1149: [CTSC2007]风玲Mobiles

1149: [CTSC2007]风玲Mobiles

Time Limit: 10 Sec   Memory Limit: 162 MB
Submit: 650   Solved: 349
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Description

   

Input

Output

输出仅包含一个整数。表示最少需要多少次交换能使风铃满足Ike的条件。如果不可能满足，输出-1。

Sample Input

6
2 3
-1 4
5 6
-1 -1
-1 -1
-1 -1

Sample Output

2

HINT

Source

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不能的情况很显然

如果深度最浅和最深的铃铛差值大于1，那就无解

因为无论怎样交换都无法改变铃铛的深度，所以如果左右交换一次仍不能符合题意，无解

#include
#include
#include
#include
#include
using namespace std;

const int maxn = 5E5 + 10;
int n,ans,mi = maxn,ma,cnt,L[maxn],L1[maxn],L2[maxn];
bool flag,bo[maxn],f[4];

vector v[maxn];

bool Judge(int x,int y) {return L1[x] >= L2[y];}
void Dfs(int x)
{
	if (bo[x]) {
		L1[x] = L2[x] = L[x];
		mi = min(mi,L[x]);
		ma = max(ma,L[x]);
		return;
	}
	L1[x] = maxn;
	for (int i = 0; i < v[x].size(); i++) {
		int to = v[x][i];
		L[to] = L[x] + 1;
		Dfs(to);
		L1[x] = min(L1[x],L1[to]);
		L2[x] = max(L2[x],L2[to]);
	}
	bool fl,fr;
	int lc = v[x][0],rc = v[x][1];
	if (Judge(lc,rc)) return;
	if (Judge(rc,lc)) {++ans; return;}
	flag = 1; 
}

int main()
{
	#ifdef DMC
		freopen("DMC.txt","r",stdin);
	#endif
	
	cin >> n; cnt = n;
	for (int i = 1; i <= n; i++) {
		int x;
		scanf("%d",&x);
		if (x == -1) x = ++cnt,bo[x] = 1;
		v[i].push_back(x);
		scanf("%d",&x);
		if (x == -1) x = ++cnt,bo[x] = 1;
		v[i].push_back(x);
	}
	L[1] = 1; Dfs(1);
	if (ma - mi > 1) {
		cout << -1;
		return 0;
	}
	if (flag) cout << -1;
	else cout << ans;
	return 0;
}