ZOJ 3795 Grouping（強聯通分量 + 縮點 + Dp)

標籤：acm   dfs   dp   

Problem Description:

Suppose there are N people in ZJU, whose ages are unknown. We have some messages about them. Thei-th message shows that the age of person si is not smaller than the age of personti. Now we need to divide all theseN 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 integersN(1≤ N≤ 100000), M(1≤ M≤ 300000), N is the number of people, and M is is the number of messages. Then followed byM lines, each line contain two integers si and ti. 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

解題思路：
強連通分量縮點，然後DP求最長路

#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];//點對應的強聯通分量的編號int dfs_clock, scc_cnt;//強聯通分量的個數，編號為1~scc_cntint num[MAXN];//每個強聯通分量所含的點的數目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（強聯通分量 + 縮點 + Dp)