圖論trainning-part-2 C. The Largest Clique

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C. The Largest CliqueTime Limit: 3000msMemory Limit: 131072KB64-bit integer IO format: %lld      Java class name: Main  

Given a directed graph G, consider the following transformation. First, create a new graph T(G) to have the same vertex set as G. Create a directed edge between two vertices u and v in T(G) if and only if there is a path between u and v in G that follows the directed edges only in the forward direction. This graph T(G) is often called the transitive closure of G.

We define a clique in a directed graph as a set of vertices U such that for any two vertices u and v in U, there is a directed edge either from u to v or from v to u (or both). The size of a clique is the number of vertices in the clique.

The number of cases is given on the first line of input. Each test case describes a graph G. It begins with a line of two integers nand m, where 0 ≤ n ≤ 1000 is the number of vertices of G and 0 ≤ m ≤ 50,000 is the number of directed edges of G. The vertices ofG are numbered from 1 to n. The following m lines contain two distinct integers u and v between 1 and n which define a directed edge from u to v in G.

For each test case, output a single integer that is the size of the largest clique in T(G).

Sample input

15 51 22 33 14 15 2

Output for sample input

4

解題：強連通子圖的求解，縮點，DAG上的動態規劃。先求出所有的強連通子圖後，再對各個強連通子圖進行縮點，所謂縮點，即為把這個強連通塊作為一個點，進行新圖的建立。原來圖上的任意一點必然屬於某個強連通塊。所以根據各點所在的連通塊，進行新圖的建立，注意方向性，DAG上的動態規劃是對於有向圖而言的，所以必須保證方向的正確性。建立新圖後，求DAG上的最長路徑即可。

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define INF 0x3f3f3f3f15 using namespace std;16 const int maxn = 1010;17 int low[maxn],dfn[maxn],iindex,sccBlocks;18 bool instack[maxn],vis[maxn];19 int belong[maxn],val[maxn],dp[maxn],n,m;20 stack<int>s;21 vector<int>g[maxn];22 vector<int>mp[maxn];23 void tarjan(int u){24     dfn[u] = low[u] = ++iindex;25     instack[u] = true;26     s.push(u);27     for(int i = 0; i < g[u].size(); i++){28         int v = g[u][i];29         if(!dfn[v]){30             tarjan(v);31             low[u] = min(low[u],low[v]);32         }else if(instack[v] && low[u] > dfn[v]) low[u] = dfn[v];33     }34     if(dfn[u] == low[u]){35         int v;36         sccBlocks++;37         do{38             v = s.top();39             s.pop();40             instack[v] = false;41             belong[v] = sccBlocks;42         }while(u != v);43     }44 }45 int dag(int u){46     if(dp[u]) return dp[u];47     else if(mp[u].size() == 0) return dp[u] = val[u];48     int ans = 0;49     for(int v = 0; v < mp[u].size(); v++){50         ans = max(ans,dag(mp[u][v]));51     }52     return dp[u] = ans+val[u];53 }54 int main(){55     int t,u,v,i,j;56     scanf("%d",&t);57     while(t--){58         scanf("%d%d",&n,&m);59         for(i = 0; i <= n; i++){60             g[i].clear();61             dfn[i] = low[i] = 0;62             instack[i] = false;63             val[i] = belong[i] = 0;64             dp[i] = 0;65             mp[i].clear();66         }67         for(i = 0; i < m; i++){68             scanf("%d%d",&u,&v);69             g[u].push_back(v);70         }71         iindex = sccBlocks = 0;72         for(i = 1; i <= n; i++)73             if(!dfn[i]) tarjan(i);74         for(u = 1; u <= n; u++){75             val[belong[u]]++;76             memset(vis,false,sizeof(vis));77             for(j = 0; j < g[u].size(); j++){78                 v = g[u][j];79                 if(!vis[belong[v]] && belong[v] != belong[u]){80                     vis[belong[v]] = true;81                     mp[belong[u]].push_back(belong[v]);82                 }83             }84         }85         int ans = 0;86         for(i = 1; i <= sccBlocks; i++)87             ans = max(ans,dag(i));88         printf("%d\n",ans);89     }90     return 0;91 }

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