ZOJ 3795 Grouping（强联通分量 + 缩点 + Dp)

标签：acm   dfs   dp   

Problem Description:

Suppose there are N people in ZJU, whose ages are unknown. We have some messages about them. The i-th message shows that the age of person s_i is not smaller than the age of person t_i. Now we need to divide all these N people into several groups. One‘s age shouldn‘t be compared with each other in the same group, directly or indirectly. And everyone should be assigned to one and only one group. The task is to calculate the minimum number of groups that meet the requirement.

Input:

There are multiple test cases. For each test case: The first line contains two integers N(1≤ N≤ 100000), M(1≤ M≤ 300000), N is the number of people, and M is is the number of messages. Then followed by M lines, each line contain two integers s_i and t_i. There is a blank line between every two cases. Process to the end of input.

Output:

For each the case, print the minimum number of groups that meet the requirement one line.

Sample Input:

4 4
1 2
1 3
2 4
3 4

Sample Output:

3

解题思路：
强连通分量缩点，然后DP求最长路

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
const int MAXN = 100000 + 10;
vector<int>G[MAXN];
vector<int>RG[MAXN];
int vis[MAXN];
int pre[MAXN], lowlink[MAXN], sccno[MAXN];//点对应的强联通分量的编号
int dfs_clock, scc_cnt;//强联通分量的个数，编号为1~scc_cnt
int num[MAXN];//每个强联通分量所含的点的数目
stack<int>S;
void dfs(int u)
{
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    for(int i=0;i<G[u].size();i++)
    {
        int v = G[u][i];
        if(!pre[v])
        {
            dfs(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if(!sccno[v])
        {
            lowlink[u] = min(lowlink[u], pre[v]);
        }
    }
    if(lowlink[u] == pre[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top();
            S.pop();
            sccno[x] = scc_cnt;
            num[scc_cnt]++;
            if(x == u) break;
        }
    }
}
void find_scc(int n)
{
    dfs_clock = scc_cnt = 0;
    memset(sccno, 0, sizeof(sccno));
    memset(pre, 0, sizeof(pre));
    memset(num, 0, sizeof(num));
    for(int i=0;i<n;i++)
    {
        if(!pre[i]) dfs(i);
    }
}
int dp[MAXN];
int DP(int u)
{
    if(dp[u] != -1)
        return dp[u];
    dp[u] = num[u];
    for(int i=0;i<RG[u].size();i++)
    {
        int v = RG[u][i];
        dp[u] = max(dp[u], num[u] + DP(v));
    }
    return dp[u];
}
int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        for(int i=0;i<n;i++)
            G[i].clear();
        int u, v;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d", &u, &v);
            u--; v--;
            G[u].push_back(v);
        }
        find_scc(n);
        for(int i=1;i<=scc_cnt;i++) RG[i].clear();
        for(int i=0;i<n;i++)
        {
            int sz = G[i].size();
            for(int j=0;j<sz;j++)
            {
                if(sccno[i] != sccno[G[i][j]])
                {
                    RG[sccno[i]].push_back(sccno[G[i][j]]);
                }
            }
        }
        int ans = 0;
        memset(dp, -1, sizeof(dp));
        for(int i=1;i<=scc_cnt;i++)
            ans = max(ans, DP(i));
        printf("%d\n", ans);
    }
    return 0;
}

ZOJ 3795 Grouping（强联通分量 + 缩点 + Dp)

标签：acm   dfs   dp   

原文地址：http://blog.csdn.net/moguxiaozhe/article/details/45134009