每日算法

每日算法

those times when you get up early and you work hard; those times when you stay up late and you work hard; those times when don’t feel like working — you’re too tired, you don’t want to push yourself — but you do it anyway. That is actually the dream. That’s the dream. It’s not the destination, it’s the journey. And if you guys can understand that, what you’ll see happen is that you won’t accomplish your dreams, your dreams won’t come true, something greater will. mamba out

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那些你早出晚归付出的刻苦努力，你不想训练，当你觉的太累了但还是要咬牙坚持的时候，那就是在追逐梦想，不要在意终点有什么，要享受路途的过程，或许你不能成就梦想，但一定会有更伟大的事情随之而来。 mamba out~

2020.3.25

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P5198 [USACO19JAN]Icy Perimeter S

蓝题就这么被AC了 这题确实有点水只需要看出来周长就好了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;
const int N = 1010;

char a[N][N];
bool vis[N][N];
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
int A = 0,S = 0 , n , ansa = 0, anss = 0;

struct node{
	int x, y;
	node(int x,int y):x(x),y(y){}
};
void bfs(int sx,int sy)
{
	queue<node> q;
	q.push(node(sx,sy));
	vis[sx][sy] = 1;
	
	while(!q.empty())
	{
		node now = q.front();
		
		for(int i = 0;i < 4; i ++)
		{
			int nx = now.x + dir[i][0];
			int ny = now.y + dir[i][1];
			if(a[nx][ny] == '.')S++;
			if(a[nx][ny] == '#' && !vis[nx][ny])
			{
				A++;
				vis[nx][ny] = 1;
				q.push(node(nx,ny));
			}
		}
		q.pop();
	}
}

void input()
{
	fill(a[0],a[0] + N * N, '.');
	cin >> n;
	for(int i = 1;i <= n ;i ++)
	{
		for(int j = 1;j <= n ;j ++)
		{
			cin >> a[i][j];
		}
	}
}

int main()
{
	input();
	for(int i = 1;i <= n ;i ++)
	{
		for(int j = 1;j <= n ;j ++)
		{
			if(!vis[i][j] && a[i][j] == '#')
			{
				A = 1;S = 0;
				bfs(i , j);
				if(A > ansa)
				{
					ansa = A;
					anss = S;
				}else if(A == ansa)
				{
					if(anss > S)anss = S;
				}
			}
		}
	}
	cout << ansa << " " << anss;
	return 0;
}

P2068 统计和

单点修改，区间查询

#include <iostream>
#include <cstdio>
#include <algorithm>
#define lowbit(x) (x & -x)
using namespace std;
const int N = 100005;
int tr[N] , a[N] , n , m;

void add(int x,int v)
{
	for(int i = x;i <= n ;i += lowbit(i))tr[i] += v;
}

long long query(int x)
{
	long long ans = 0;
	for(int i = x;i;i-=lowbit(i))
		ans += tr[i];
	return ans;
}

int main()
{
	cin >> n >> m;
	for(int i = 1;i <= n ;i ++)
	{
		scanf("%d",&a[i]);
		add(i,a[i]);	
	}
	for(int i = 0;i < m ;i ++)
	{
		char ch;int a, b ;
		cin >> ch;
		scanf("%d %d",&a,&b);
		if(ch == 'x'){
			add(a,b);
		}else{
			printf("%lld
",query(b)-query(a-1));
		}
	}
	return 0;
} 

P2360 地下城主

原题是POJ的题目 被搬过来 做法差不多就是变成一次性的数据了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>

using namespace std;
const int N = 105;


char a[N][N][N];
bool vis[N][N][N];
int l , r, c;
int sl, sx ,sy;
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
int updown[2] = {1,-1};
struct node{
	int l , x, y , time;
};

bool inmap(int ll ,int x,int y)
{
	return ll >= 1 && ll <= l && x >= 1 && x <= r && y >= 1 && y <= c;	
}

bool check(int l , int x , int y)
{
	if(inmap(l,x,y) && !vis[l][x][y] && a[l][x][y] != '#')
	{
		return true;
	}else return false;
}

int bfs()
{
	vis[sl][sx][sy] = 1;
	queue<node> q;
	q.push({sl, sx ,sy , 0});
	bool flag = 0;
	while(!q.empty())
	{
		node now = q.front();
		// 在当前层不变
		 
		for(int i = 0;i < 4;i ++)
		{
			int nx = now.x + dir[i][0];
			int ny = now.y + dir[i][1];
			int nl = now.l;
			if(check(nl,nx,ny))
			{
				if(a[nl][nx][ny] == 'E')
				{
					flag = 1;
					return now.time + 1;
				}
				vis[nl][nx][ny] = 1;
				q.push({nl,nx,ny,now.time + 1});	
			}	
		}
		// 值改变当前层数
		for(int i = 0;i < 2;i ++)
		{
			int nx = now.x,ny = now.y;
			int nl = now.l + updown[i];
			if(check(nl,nx,ny))
			{
				if(a[nl][nx][ny] == 'E')
				{
					flag = 1;
					return now.time + 1;
				}
				vis[nl][nx][ny] = 1;
				q.push({nl,nx,ny,now.time + 1});	
			}	
		} 
		q.pop();
	}
	if(!flag) return -1;
}

void output(int ans)
{
	if(ans == -1)
	{
		cout << "Trapped!" << endl;
		return;	
	}else{
		printf("Escaped in %d minute(s).
",ans);
	}
}

void input()
{
	string s;
	for(int k = 1;k <= l;k ++)
	{
		for(int i = 1;i <= r ;i ++)
		{
			cin >> s;
			for(int j = 0;j < s.size() ;j ++)
			{
				a[k][i][j+1] = s[j];
				if(a[k][i][j+1] == 'S')
				{
					sl = k;sx = i;sy = j + 1;		
				}
			}
		}	
	}	
}
void clear()
{
	for(int k = 1;k <= l;k ++)
	{
		for(int i = 1;i <= r ;i ++)
		{
			for(int j = 1;j <= c ;j ++)
			{
				vis[k][i][j] = 0;
			}
		}	
	}
}
int main()
{
	cin >> l >> r >> c;
	input();
	int dis = bfs();
	output(dis);
	clear();
	return 0;
}

CF580C Kefa and Park

嗯 这最后搜的时候其实是判断是否到达最后一个还满足条件

#include<bits/stdc++.h>

using namespace std;
const int N = 1000005;

int arr[N] , n , m ;
long long ans = 0;
vector<int> a[N];
bool vis[N];

void dfs(int now,int val)
{
	if(val > m)return;
	int t = 1;
	vis[now] = 1;
	for(int i = 0;i < a[now].size();i++)
	{
		if(!vis[a[now][i]])
		{
			t = 0;	
			if(!arr[a[now][i]])
			{
				dfs(a[now][i],0);
			}else dfs(a[now][i],val + 1);
		}
	}
	ans += t;
	return;
}

void input()
{
	cin >> n >> m;
	for(int i = 1;i <= n;i ++)
		scanf("%d",&arr[i]);
	for(int i = 0;i < n - 1;i ++)
	{
		int x , y;
		scanf("%d %d",&x , &y);
		a[x].push_back(y);
		a[y].push_back(x); 
	}
}
int main()
{
	input();
	dfs(1,arr[1]);
	cout << ans << endl;
	return 0;
}