HDOJ 3639 Hawk-and-Chicken

最新推荐文章于 2017-09-29 16:01:57 发布

最新推荐文章于 2017-09-29 16:01:57 发布 · 369 阅读

强连通及其拓展专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种解决幼儿园游戏中角色分配问题的算法。通过投票和传递支持的方式确定扮演鹰的角色，采用图论中的强连通分量和拓扑排序思想实现算法。提供了两种实现方案：深度优先搜索(DFS)和广度优先搜索(BFS)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Hawk-and-Chicken

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3081 Accepted Submission(s): 954

Problem Description

Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But there always exists a big problem: every kid in this game want to play the role of Hawk.
So the teacher came up with an idea: Vote. Every child have some nice handkerchiefs, and if he/she think someone is suitable for the role of Hawk, he/she gives a handkerchief to this kid, which means this kid who is given the handkerchief win the support. Note the support can be transmitted. Kids who get the most supports win in the vote and able to play the role of Hawk.(A note:if A can win
support from B(A != B) A can win only one support from B in any case the number of the supports transmitted from B to A are many. And A can't win the support from himself in any case.
If two or more kids own the same number of support from others, we treat all of them as winner.
Here's a sample: 3 kids A, B and C, A gives a handkerchief to B, B gives a handkerchief to C, so C wins 2 supports and he is choosen to be the Hawk.

Input

There are several test cases. First is a integer T(T <= 50), means the number of test cases.
Each test case start with two integer n, m in a line (2 <= n <= 5000, 0 <m <= 30000). n means there are n children(numbered from 0 to n - 1). Each of the following m lines contains two integers A and B(A != B) denoting that the child numbered A give a handkerchief to B.

Output

For each test case, the output should first contain one line with "Case x:", here x means the case number start from 1. Followed by one number which is the total supports the winner(s) get.
Then follow a line contain all the Hawks' number. The numbers must be listed in increasing order and separated by single spaces.

Sample Input

Sample Output

Case 1: 2
0 1
Case 2: 2
0 1 2

一：DFS

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset(a,(b),sizeof(a))
#define f(i,a,b) for(int i=a;i<b;i++)
#define maxn 5005
#define INF 2e9;
int low[maxn],dfn[maxn],in[maxn],vis[maxn];
int instack[maxn],point[maxn],ans[maxn],dp[maxn];
int n,m,tot,num,sum;
vector<int>G1[maxn],G2[maxn];
stack<int>S;
void init()
{
    tot=num=0;
    f(i,0,n)
    {
        G1[i].clear();
        G2[i].clear();
        low[i]=dfn[i]=in[i]=vis[i]=0;
        instack[i]=point[i]=ans[i]=dp[i]=0;
    }
    while(!S.empty())
    {
        S.pop();
    }
}
void tarjan(int x)
{
    low[x]=dfn[x]=tot++;
    vis[x]=instack[x]=1;
    S.push(x);
    f(i,0,G1[x].size())
    {
        int v=G1[x][i];
        if(!vis[v])
        {
            tarjan(v);
            low[x]=min(low[x],low[v]);
        }
        else if(instack[v])
        {
            low[x]=min(low[x],dfn[v]);
        }
    }
    if(low[x]==dfn[x])
    {
        int cnt=0;
        while(1)
        {
            int v=S.top();
            S.pop();
            instack[v]=0;
            cnt++;
            point[v]=num;
            if(v==x) break;
        }
        ans[num]=cnt;
        num++;
    }
}
void dfs(int x)
{
    vis[x]=1;
    sum+=ans[x];
    f(i,0,G2[x].size())
    {
        if(!vis[G2[x][i]])
            dfs(G2[x][i]);
    }
}
int main()
{
    int T,t=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        int x,y;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            G1[x].push_back(y);
        }
        f(i,0,n)
        {
            if(!vis[i])
                tarjan(i);
        }
        f(i,0,n)
        f(j,0,G1[i].size())
        {
            if(point[i]!=point[G1[i][j]])
            {
                in[point[i]]++;
                G2[point[G1[i][j]]].push_back(point[i]);
            }
        }
        int res=-1;
        f(i,0,num)
        {
            if(!in[i])
            {
                sum=0;
                mst(vis,0);
                dfs(i);
                dp[i]=sum;
                res=max(res,sum);
            }
        }
        int flag=1;
        printf("Case %d: %d\n",t++,res-1);
        f(i,0,n)
        {
            if(dp[point[i]]==res)
            {
                if(flag)
                {
                    printf("%d",i);
                    flag=0;
                }
                else printf(" %d",i);
            }
        }
        printf("\n");
    }
    return 0;
}

二：BFS

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset(a,(b),sizeof(a))
#define f(i,a,b) for(int i=a;i<b;i++)
#define maxn 5005
#define INF 2e9;
int low[maxn],dfn[maxn],in[maxn],vis[maxn];
int instack[maxn],point[maxn],ans[maxn],dp[maxn];
int n,m,tot,num,sum;
vector<int>G1[maxn],G2[maxn];
stack<int>S;
void init()
{
    tot=num=0;
    f(i,0,n)
    {
        G1[i].clear();
        G2[i].clear();
        low[i]=dfn[i]=in[i]=vis[i]=0;
        instack[i]=point[i]=ans[i]=dp[i]=0;
    }
    while(!S.empty())
    {
        S.pop();
    }
}
void tarjan(int x)
{
    low[x]=dfn[x]=tot++;
    vis[x]=instack[x]=1;
    S.push(x);
    f(i,0,G1[x].size())
    {
        int v=G1[x][i];
        if(!vis[v])
        {
            tarjan(v);
            low[x]=min(low[x],low[v]);
        }
        else if(instack[v])
        {
            low[x]=min(low[x],dfn[v]);
        }
    }
    if(low[x]==dfn[x])
    {
        int cnt=0;
        while(1)
        {
            int v=S.top();
            S.pop();
            instack[v]=0;
            cnt++;
            point[v]=num;
            if(v==x) break;
        }
        ans[num]=cnt;
        num++;
    }
}
void bfs(int x)
{
    queue<int>q;
    vis[x]=1;
    q.push(x);
    while(!q.empty())
    {
        int v=q.front();
        q.pop();
        sum+=ans[v];
        f(i,0,G2[v].size())
        {
            if(!vis[G2[v][i]])
            {
                vis[G2[v][i]]=1;
                q.push(G2[v][i]);
            }
        }
    }
}
int main()
{
    int T,t=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        int x,y;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            G1[x].push_back(y);
        }
        f(i,0,n)
        {
            if(!vis[i])
                tarjan(i);
        }
        f(i,0,n)
        f(j,0,G1[i].size())
        {
            if(point[i]!=point[G1[i][j]])
            {
                in[point[i]]++;
                G2[point[G1[i][j]]].push_back(point[i]);
            }
        }
        int res=-1;
        f(i,0,num)
        {
            if(!in[i])
            {
                sum=0;
                mst(vis,0);
                bfs(i);
                dp[i]=sum;
                res=max(res,sum);
            }
        }
        int flag=1;
        printf("Case %d: %d\n",t++,res-1);
        f(i,0,n)
        {
            if(dp[point[i]]==res)
            {
                if(flag)
                {
                    printf("%d",i);
                    flag=0;
                }
                else printf(" %d",i);
            }
        }
        printf("\n");
    }
    return 0;
}