Problem B: The Largest Clique
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 n and 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 of G 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
Zachary Friggstad
题目大意：
T组测试数据，给一张有向图G，求一个结点数最大的结点集，使得该结点中任意两个结点 u 和 v满足：要么 u 可以到达 v， 要么 v 可以到达 u（u 和 v 相互可达也可以）。
解题思路：
”同一个强连通分量中的点要么都选，要么不选。把强连通分量收缩点后得到SCC图，让每个SCC结点的权等于它的结点数，则题目转化为求SCC图上权最大的路径。由于SCC图是一个 DAG， 可以用动态规划求解。“
解题代码：
#include
#include
#include
#include
#include
using namespace std;
const int maxn=1100;
const int maxm=51000;
struct edge{
int u,v,next;
edge(int u0=0,int v0=0){
u=u0;v=v0;
}
}e[maxm];
int n,m,head[maxn],dfn[maxn],low[maxn],mark[maxn],w[maxn],color[maxn],dp[maxn],cnt,nc,index;
vector vec;
vector > dfsmap;
void addedge(int u,int v){
e[cnt]=edge(u,v);e[cnt].next=head[u];head[u]=cnt++;
}
void input(){
cnt=nc=index=0;
scanf("%d%d",&n,&m);
vec.clear();
for(int i=0;i<=n;i++){
w[i]=dfn[i]=0;
mark[i]=false;
color[i]=dp[i]=head[i]=-1;
}
int u,v;
while(m-- >0){
scanf("%d%d",&u,&v);
addedge(u,v);
}
}
void tarjan(int s){
dfn[s]=low[s]=++index;
mark[s]=true;
vec.push_back(s);
for(int i=head[s];i!=-1;i=e[i].next){
int d=e[i].v;
if(!dfn[d]){
tarjan(d);
low[s]=min(low[d],low[s]);
}else if(mark[d]){
low[s]=min(low[s],dfn[d]);
}
}
if(dfn[s]==low[s]){
nc++;
int d;
do{
d=vec.back();
vec.pop_back();
color[d]=nc;
mark[d]=false;
w[nc]++;
}while(d!=s);
}
}
int DP(int s){
if(dp[s]!=-1) return dp[s];
int ans=w[s];
for(int i=0;ians) ans=DP(d)+w[s];
}
return dp[s]=ans;
}
void solve(){
for(int i=1;i<=n;i++){
if(!dfn[i]) tarjan(i);
}
dfsmap.clear();
dfsmap.resize(nc+1);
for(int i=0;i"<ans) ans=DP(i);
//cout<0){
input();
solve();
}
return 0;
}