图论trainning-part-2 C. The Largest Clique

C. The Largest Clique

Time Limit: 3000ms

Memory Limit: 131072KB

64-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

1

5 5

1 2

2 3

3 1

4 1

5 2

Output for sample input

4

解题：强连通子图的求解，缩点，DAG上的动态规划。先求出所有的强连通子图后，再对各个强连通子图进行缩点，所谓缩点，即为把这个强连通块作为一个点，进行新图的建立。原来图上的任意一点必然属于某个强连通块。所以根据各点所在的连通块，进行新图的建立，注意方向性，DAG上的动态规划是对于有向图而言的，所以必须保证方向的正确性。建立新图后，求DAG上的最长路径即可。

 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 long

14 #define INF 0x3f3f3f3f

15 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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