Daliy Algorithm (GPLT)-- day 94

Nothing to fear

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种一棵树最好的时间是十年前，其次是现在！

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~

2020.7.6 - 7.7

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人一我十，人十我百，追逐青春的梦想，怀着自信的心，永不言弃！

GPLT - L2-020 功夫传人

坑

• 只统计得道者得功夫值！读题要仔细
• 祖师爷也可能是得道者
• printf("%.f",ans) 会自动进行四舍五入，而题目要求只输出整数部分，故只能利用强制类型转换。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <vector>

using namespace std;
const int N = 100005;
int n;
double z , r;
vector<int> a[N];
double tot;
int dedao[N];
void dfs(int now,double val)
{
	if(dedao[now]){
		tot += val * dedao[now];
	}
	for(int i = 0;i < a[now].size(); i++)
	{
		dfs(a[now][i] ,val - val * r * 0.01);
	}
}
int main()
{
	cin >> n >> z >> r;
	int k = 0;
	for(int i = 0;i < n ;i ++)
	{
		cin >> k;
		if(k == 0){
			scanf("%d",&dedao[i]);
		}else{
			int x = 0;
			for(int j = 0;j < k;j ++)
			{
				scanf("%d", &x);
				a[i].push_back(x);
			}
		}
	}	
	dfs(0 , z);
	printf("%d
",(int)tot);
	return 0;	
}

GPLT - L2-021 点赞狂魔

分析：

利用set记录不同得标签出现得次数，标签得平均出现次数为总数量 / 不同标签得数目。然后按照题目规则进行排序即可。

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;

const int N = 505;

struct node{
	string name;
	int cnt;
	int k;
}a[N];

bool cmp(node a ,node b)
{
	if(a.cnt == b.cnt)
	{
		return (a.k * 1.0 / a.cnt) < (b.k * 1.0 / b.cnt);
	}else return a.cnt > b.cnt;
}
int main()
{
	int n;
	cin >> n;
	set<int> s;
	for(int i = 0;i < n ;i ++)
	{
		cin >> a[i].name;
		cin >> a[i].k;
		for(int j = 0;j < a[i].k; j++)
		{
			int x = 0;
			scanf("%d" ,&x);
			s.insert(x);
		}
		a[i].cnt = s.size();
		s.clear();
	}
	sort(a , a + n , cmp);
	for(int i = 0;i < 3;i ++)
	{
		if(a[i].k!=0)
		{
			if(i == 0)
				cout << a[i].name;
			else cout << " " << a[i].name;
		}else{
			if(i == 0)
				cout << "-";
			else cout << " -"; 
		}
	}
	return 0;
}

GPLT - L2-022 重排链表

没认真读题还以为要排序呢。。。，注意链表去重，老玩家了

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

using namespace std;

const int N = 1000005;
struct node{
	int add , val , next;
}a[N] , Link[N] ,ans[N];
int head , n , len = 0;
bool cmp(node a,node b)
{
	return a.val < b.val;
}
int main()
{
	cin >> head >> n;
	int x , y , z;
	for(int i = 0;i < n;i ++)
	{
		scanf("%d %d %d",&x , & y, &z);
		a[x] = {x , y , z};
	}
	for(int i = head ;i != -1 ;i = a[i].next)
	{
		Link[len++] = a[i];
	}
	int i = 0 , j = len - 1 ,id = 0;
	while(i < j)
	{
		ans[id++] = Link[j--];
		ans[id++] = Link[i++];
		if(i == j){
			ans[id++] = Link[i];
			break;
		}
	}
	for(int i = 0;i < len ;i ++)
	{
		if(i == len - 1)
		{
			printf("%05d %d -1
",ans[i].add , ans[i].val);
		}else{
			printf("%05d %d %05d
",ans[i].add , ans[i].val ,ans[i+1].add);
		}
	}
	return 0;
}

GPLT - L2-023 图着色问题

PAT叫我做人

题目只说了图是合法得，没说图是一定联通得，所以要遍历每一个未被访问得结点，不然数据点3是过不去得 惨遭毒手

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <cstring>
using namespace std;

const int N = 250005;
const int M = 505;
struct edge{
	int next;
	int to;
}e[N];
int head[M],tot;
int  n, m, k;
int color[M];
bool vis[M];
void add(int a,int b)
{
	e[tot].to = b;
	e[tot].next = head[a];
	head[a] = tot++;
}
set<int> s;
void checkColor(int now, bool &flag)
{
	vis[now] = true;
	s.insert(color[now]);
	for(int i = head[now] ; i ; i = e[i].next)
	{
		int next = e[i].to;
		if(color[now] == color[next]){
			flag = true;
			return;
		}
		if(!vis[next] && !flag)
		{
			checkColor(next , flag);
		}
	}
}
int main()
{
	cin >> n >> m >> k;
	int x , y;
	for(int i = 0;i < m;i ++)
	{
		scanf("%d %d",&x , &y);
		add(x , y);add(y , x);
	}
	int q;
	cin >> q;
	for(int i = 0;i < q;i ++)
	{
		for(int j = 1;j <= n;j ++){
			scanf("%d",&color[j]);
		}
		memset(vis , 0 , sizeof vis);
		bool flag = false;
		s.clear();
		for(int i = 1;i <= n ;i ++)
		{
			if(!vis[i] && !flag)
			{
				checkColor(i, flag);
			}
		}
		if(!flag && s.size() == k)printf("Yes
");
		else printf("No
");
	}
	return 0;
}

GPLT - L2-024 部落

并查集模板题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>


using namespace std;

const int N = 10005;
int f[N] , cnt[N];
bool vis[N];
int find(int k)
{
	if(k == f[k])return k;
	else return f[k] = find(f[k]);
}
int Unoin(int a,int b)
{
	int x = find(a) , y = find(b);
	f[x] = y;
}
int main()
{
	for(int i = 1;i <= N;i ++)f[i] = i;
	int n , q;
	cin >> n;
	for(int i = 0;i < n;i ++)
	{
		int k = 0;cin >> k;
		for(int j = 0;j < k;j ++)
		{
			cin >> cnt[j];
			vis[cnt[j]] = 1;
		}
		for(int j = 1;j < k;j ++)
		Unoin(cnt[j-1],cnt[j]);
	}
	int people = 0 , ans = 0;
	for(int i = 1;i <= N;i ++)
	{
		if(vis[i] == 1){
			if(i == f[i])ans++;
			people++;
		}
	}
	cout << people << " " << ans << endl;
	cin >> q;
	int a , b;
	for(int i = 0;i < q;i ++)
	{
		cin >> a >> b;
		int x = find(a) , y = find(b);
		if(x != y)printf("N
");
		else printf("Y
");
	}
	return 0;
}

GPLT - L2-025 分而治之

简单dfs判断连通分量即可，话说 说好的m不超过10000呢

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <cstring>
using namespace std;

const int N = 100005;
struct edge{
	int to;
	int next;
}e[N];
int head[N] , tot;
int n , m;
bool vis[N];

void add(int a , int b)
{
	e[tot].to = b;
	e[tot].next = head[a];
	head[a] = tot++;
}

bool dfs(int x)
{
	for(int i = head[x] ; i ; i = e[i].next)
	{
		int next = e[i].to;
		if(vis[next] == false)return false;
	}
	return true;
}
int main()
{
	cin >> n >> m;
	int x , y;
	for(int i = 0;i < m;i ++)
	{
		cin >> x >> y;
		if(x == y)continue;
		add(x , y) ,add(y , x);
	}
	int k , np;
	cin >> k;
	for(int time = 0;time < k;time ++)
	{
		cin >> np;
		memset(vis , 0 , sizeof vis);
		for(int i = 0;i < np;i ++)
		{
			cin >> x;
			vis[x] = 1;
		}
		bool flag = true;
		for(int i = 1;i <= n;i ++)
		{
			if(!vis[i])
			{
				if(!dfs(i))
				{
					flag = false;
					break;
				}
			}
		}
		if(flag)printf("YES
");
		else printf("NO
");
	}
	return 0;
}