Code Force 645 C

题目链接：Enduring Exodus

比赛的时候想着用 bfs 来做的，第十个点一直超时！！！很难受了

然后比赛完了才知道这是二分答案的题！

好吧，再补充补充知识点！

#include<iostream>
#include<cstdio>
using namespace std;
int a[100100];
int pre[100100]; // 前缀和，方便计算
int n,m;
bool check(int k)
{
	int sum = 0;
	for(int i=1;i<=n;i++)
	{
		if(a[i] != 1)
		{
			int l = max(i-k,1);  
        	        int r = min(i+k,n);  
			if(pre[r] - pre[l-1]-1 >= m)
				return true;
		}	
	}
	return false;
}
int main()
{
	scanf("%d%d",&n,&m);
	getchar();
	for(int i=1;i<=n;i++)
	{
		char ch = getchar();
		if(ch == '1')
		{
			a[i] = 1;
			pre[i] = pre[i-1];
		}
		else {
			a[i] = 0;
			pre[i] = pre[i-1] + 1;
		}
	}
	int l = 0,r = n,ans = 1e9;
	while(l + 1 < r)
	{
		int mid = ( l + r) >> 1;
		if(check(mid))
		{
			r = mid;
			ans = mid;
		} 
		else l = mid;
	}
	printf("%d\n",ans);
	return 0;
}

超时代码如下：

#include<iostream>
#include<cstdio>
#include<queue> 
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 100100;
bool vis[N];
bool viss[N];
int n,m,ans;
queue<int>q;
int minn = N;
void bfs(int k)
{
	while(ans < m)
	{
		int pos = q.front();
			q.pop();
		viss[pos] = 1;
		if(vis[pos] == 0 && pos!=k)
		{
			ans ++;
			if(ans == m)
			{
				minn= min(minn,abs(k-pos));
				return ;
			}
		}
		if( pos-1 >= 1 && viss[pos-1] == 0)
		{
			viss[pos-1] = 1;
			q.push(pos-1);
		}
		if(pos+1<=n&&viss[pos+1] == 0)
		{
			viss[pos+1] = 1;
			q.push(pos+1);
		}
	}
}
int main()
{
	//while(scanf("%d%d",&n,&m)!=EOF)
	//{
		scanf("%d%d",&n,&m);
		minn = N;
		getchar();
		for(int i=1;i<=n;i++)
		{
			char ch;
			//cin>>ch;
			ch = getchar();
			if(ch == '1')
			vis[i] = 1;
			else vis[i] = 0;
		}
		for(int i=1;i<=n;i++)
		{
			if(vis[i] == 0)
			{
				memset(viss,0,sizeof(viss));
				while(!q.empty())
				q.pop();
				ans = 0;
				viss[i] = 1;
				q.push(i);
				bfs(i);
			}
		}
		if(minn<=0)
		printf("%d\n",0-minn);
		else printf("%d\n",minn);
	//}
	return 0;
}

今天看博客，发现二分答案体型很固定！ 

最大值最小化，最小值最大化！

考察的是 check 函数的写法！