列表

详情

NC24159. [USACO 2015 Jan G]Grass Cownoisseur

描述

In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way cow path connecting a pair of fields. For example, if a path connects from field X to field Y, then cows are allowed to travel from X to Y but not from Y to X. Bessie the cow, as we all know, enjoys eating grass from as many fields as possible. She always starts in field 1 at the beginning of the day and visits a sequence of fields, returning to field 1 at the end of the day. She tries to maximize the number of distinct fields along her route, since she gets to eat the grass in each one (if she visits a field multiple times, she only eats the grass there once). As one might imagine, Bessie is not particularly happy about the one-way restriction on FJ's paths, since this will likely reduce the number of distinct fields she can possibly visit along her daily route. She wonders how much grass she will be able to eat if she breaks the rules and follows up to one path in the wrong direction. Please compute the maximum number of distinct fields she can visit along a route starting and ending at field 1, where she can follow up to one path along the route in the wrong direction. Bessie can only travel backwards at most once in her journey. In particular, she cannot even take the same path backwards twice.

输入描述

The first line of input contains N and M, giving the number of fields
and the number of one-way paths (1 <= N, M <= 100,000).
The following M lines each describe a one-way cow path.  Each line
contains two distinct field numbers X and Y, corresponding to a cow
path from X to Y.  The same cow path will never appear more than once.

输出描述

A single line indicating the maximum number of distinct fields Bessie
can visit along a route starting and ending at field 1, given that she can
follow at most one path along this route in the wrong direction.

示例1

输入: 

7 10
1 2
3 1
2 5
2 4
3 7
3 5
3 6
6 5
7 2
4 7

输出: 

6

说明:

Here is an ASCII drawing of the sample input:
v---3-->6
7 |\ |
^\ v \ |
| \ 1 \|
| \| v
| v 5
4<--2---^
Bessie can visit pastures 1, 2, 4, 7, 2, 5, 3, 1 by traveling
backwards on the path between 5 and 3. When she arrives at 3 she
cannot reach 6 without following another backwards path.

原站题解

上次编辑到这里，代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 132ms, 内存消耗: 13588K, 提交时间: 2020-01-16 22:01:20 

#include<bits/stdc++.h>
using namespace std;
const int MX=1e5+10;
int n,m,cnt;
int dfn[MX],low[MX];
int color[MX],color_num[MX];
vector<int>ve[MX],ve1[MX],ve2[MX];
bool vis[MX],f[MX];
int sum=0;
stack<int>st;
int ans[MX];
int max_op=0;

void tarjan(int u)
{
	dfn[u]=low[u]=++cnt;
	st.push(u);
	vis[u]=1;
	for(int v:ve[u])
	{
		if(!dfn[v])
		{
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}
		else if(vis[v]) low[u]=min(low[u],dfn[v]);
	}
	int k;
	if(low[u]==dfn[u])
	{
		sum++;
		do{
			k=st.top();st.pop();
			color[k]=sum;
			color_num[sum]++;
			vis[k]=0;
		}while(k!=u&&!st.empty());
	}
}
void bfs1(int x)
{
	queue<int>qu;
	qu.push(x);
	f[x]=1;
	while(!qu.empty())
	{
		int u=qu.front();qu.pop();
		
		for(int v:ve1[u])
		{
			if(ans[u]+color_num[v]>ans[v])
			{
				ans[v]=ans[u]+color_num[v];
				f[v]=1;
				qu.push(v);
			}
		}
	}
}
int check(int x)
{
	int S=0;
	for(int i:ve1[x])
	{
		if(f[i]) S=max(S,ans[i]);
	}
	return S;
}
void bfs2(int x)
{
	queue<int>q;
	q.push(x);
	while(!q.empty())
	{
		int u=q.front();q.pop();
		for(int v:ve2[u])	
		{
			if(ans[v]<ans[u]+color_num[v])
			{
				ans[v]=ans[u]+color_num[v];
				int K=check(v);
				if(K)
				max_op=max(max_op,ans[v]+K);
				q.push(v);
			}
		}
	}
}
int main()
{
//	freopen("123.in","r",stdin); //1952
	cin>>n>>m;
	for(int i=0,x,y;i<m;i++)
	{
		cin>>x>>y;
		ve[x].push_back(y);
	}
	for(int i=1;i<=n;i++)
	{
		if(!dfn[i]) tarjan(i);
	}
	for(int i=1;i<=n;i++)
	{
		for(int v:ve[i])
		{
			if(color[i]!=color[v])
			{
				ve2[color[i]].push_back(color[v]);
				ve1[color[v]].push_back(color[i]);
			}
		}
	}
	
	ans[color[1]]=color_num[color[1]];
	bfs1(color[1]);
	for(int i:ve1[color[1]]) 
	{	
		max_op=max(ans[i],max_op);
	}
	max_op+=color_num[color[1]];
	bfs2(color[1]);
	cout<<max_op-color_num[color[1]]<<'\n';
	return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 62ms, 内存消耗: 13456K, 提交时间: 2020-09-19 11:04:24 

#include<bits/stdc++.h>
using namespace std;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
const int maxn=5e5+10;
struct E{
	int next,to,val;
}tu[maxn];
int head[maxn],cnt;
void add(int next,int to){
	tu[cnt].next=head[next];
	tu[cnt].to=to;
	head[next]=cnt++;
}
stack<int>s;
int n,m;
int dfn[maxn],low[maxn],tim,num,bel[maxn],val[maxn],vis[maxn];
void tarjan(int x){
	dfn[x]=low[x]=++tim;
	s.push(x);
	vis[x]=1;
	for(int i=head[x];~i;i=tu[i].next){
		int u=tu[i].to;
		if(!dfn[u]){
			tarjan(u);
			low[x]=min(low[x],low[u]);
		}
		else if(vis[u])low[x]=min(low[x],dfn[u]);
	}
	if(dfn[x]==low[x]){
		int j;
		num++;
		do{
			j=s.top();
			s.pop();
			vis[j]=0;
			bel[j]=num;
			val[num]++;
		}while(j!=x);
	}
}
struct data{
	int u,v;
}G[maxn];
queue<int>q;
int dis[maxn];
int main(){
	memset(head,-1,sizeof(head));
	n=read(),m=read();
	for(int i=1;i<=m;i++){
		int x=read(),y=read();
		G[i].u=x,G[i].v=y;
		add(x,y);
		add(y,x+n);
		add(x+n,y+n);
	}
	for(int i=1;i<=2*n;i++){
		if(!dfn[i])tarjan(i);
	}
	memset(tu,0,sizeof(tu));
	memset(head,-1,sizeof(head));
	cnt=0;
	for(int i=1;i<=m;i++){
		int x=G[i].u,y=G[i].v;
		if(bel[x]==bel[y])continue;
		add(bel[x],bel[y]);
		add(bel[y],bel[x+n]);
		add(bel[x+n],bel[y+n]);
	}
	memset(vis,0,sizeof(vis));
	q.push(bel[1]);
	vis[bel[1]]=1;
	dis[bel[1]]=0;
	while(!q.empty()){
		int x=q.front();
		q.pop();
		vis[x]=0;
		for(int i=head[x];~i;i=tu[i].next){
			int u=tu[i].to;
			if(dis[u]<dis[x]+val[u]){
				dis[u]=dis[x]+val[u];
				if(!vis[u]){
					q.push(u);
					vis[u]=1;
				}
			}
		}
	}
	cout<<max(dis[bel[1]],dis[bel[n+1]]);
	return 0;
}

上一题

© 2025 nowcoder.com