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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 person ti. 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.
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 Mlines, each line contain two integers si and ti. There is a blank line between every two cases. Process to the end of input.
For each the case, print the minimum number of groups that meet the requirement one line.
4 4 1 2 1 3 2 4 3 4
3
set1= {1}, set2= {2, 3}, set3= {4}
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 100010; 4 struct arc{ 5 int to,next; 6 arc(int x = 0,int y = -1){ 7 to = x; 8 next = y; 9 } 10 }e[500000]; 11 int head[maxn],dfn[maxn],low[maxn],belong[maxn],num[maxn]; 12 bool instack[maxn]; 13 int tot,idx,scc,n,m,dp[maxn]; 14 stack<int>stk; 15 vector<int>g[maxn]; 16 void add(int u,int v){ 17 e[tot] = arc(v,head[u]); 18 head[u] = tot++; 19 } 20 void init(){ 21 for(int i = 0; i < maxn; ++i){ 22 dp[i] = head[i] = -1; 23 dfn[i] = low[i] = 0; 24 belong[i] = num[i] = 0; 25 instack[i] = false; 26 g[i].clear(); 27 } 28 idx = scc = tot = 0; 29 while(!stk.empty()) stk.pop(); 30 } 31 void tarjan(int u){ 32 dfn[u] = low[u] = ++idx; 33 instack[u] = true; 34 stk.push(u); 35 for(int i = head[u]; ~i; i = e[i].next){ 36 if(!dfn[e[i].to]){ 37 tarjan(e[i].to); 38 low[u] = min(low[u],low[e[i].to]); 39 }else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]); 40 } 41 if(low[u] == dfn[u]){ 42 scc++; 43 int v; 44 do{ 45 instack[v = stk.top()] = false; 46 belong[v] = scc; 47 num[scc]++; 48 stk.pop(); 49 }while(v != u); 50 } 51 } 52 int dfs(int u){ 53 if(dp[u] != -1) return dp[u]; 54 dp[u] = num[u]; 55 for(int i = g[u].size()-1; i >= 0; --i) 56 dp[u] = max(dp[u],num[u] + dfs(g[u][i])); 57 return dp[u]; 58 } 59 int main(){ 60 int u,v; 61 while(~scanf("%d %d",&n,&m)){ 62 init(); 63 for(int i = 0; i < m; ++i){ 64 scanf("%d %d",&u,&v); 65 add(u,v); 66 } 67 for(int i = 1; i <= n; ++i) 68 if(!dfn[i]) tarjan(i); 69 for(int i = 1; i <= n; ++i){ 70 for(int j = head[i]; ~j; j = e[j].next){ 71 if(belong[i] != belong[e[j].to]) 72 g[belong[e[j].to]].push_back(belong[i]); 73 } 74 } 75 int ret = 0; 76 for(int i = 1; i <= scc; ++i) 77 ret = max(ret,dfs(i)); 78 printf("%d\n",ret); 79 } 80 return 0; 81 }
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原文地址：http://www.cnblogs.com/crackpotisback/p/4471628.html
