HDU 3861.The King's Problem strong Unicom component + minimum path coverage

The King ' s problem

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 2947 Accepted Submission (s): 1049

problem DescriptionThe Kingdom of Silence, the king has a new problem. There is N cities in the kingdom and there is M directional roads between the cities. That means that if there are a road from U to V, and you can only go from the city u to City v, and can ' t go from the City v to the city U. In order to rule he kingdom more effectively, the king want to divide he kingdom into several states, and all city mus T belong to exactly one state.What's more , for each pair of the city (U, v), if there is one-to-go from U-V and go from V to U, (U, v) has to be Long to a same state. And the king must insure that in each state we can ether go from u to V or go from V to u between every pair of cities (U, V) without passing any city which belongs to other state.
Now the king asks-your help, he wants to know the least number of states he has to divide the kingdom into.

InputThe first line contains a single integer T, the number of test cases. And then followed T cases.

The first line for each case contains integers n, m (0 < n <= 5000,0 <= m <= 100000), the number of cities And roads in the kingdom. The next m lines each contains the integers u and V (1 <= u, v <= N), indicating that there was a road going from CIT Y u to City v.

OutputThe output should contain T lines. For each test case, you should just output a integer which is the least number of states the king has to divide into.

Sample Input13 21 21 3

Sample Output2

SourceMulti-university Training Contest 3-host by BITtitle Link: http://acm.hdu.edu.cn/showproblem.php?pid=3861Test Instructions: There are n cities, and M-strips have direction paths. Now to build some states, each city belongs to a state, if two cities u,v can reach each other, then u,v belong to the same state. If U,v is in the same state, then you can reach V or V to reach U and not go through the cities of other states. The minimum number of States to be built. idea: Because cities that can reach each other belong to the same state, the Tarjan are reduced. The new diagram created is a dag. In a graph, find the fewest paths so that the paths pass through all the points, and each path passes through a different point. This is a minimal path coverage problem. A minimal path overlay to a acyclic graph (DAG)

Code:

#include <iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<stack>#include<vector>#include<Set>using namespacestd;#definePI ACOs (-1.0)typedefLong LongLl;typedef pair<int,int>P;Const intmaxn=1e4+ -, maxm=1e5+ -, inf=0x3f3f3f3f, mod=1e9+7;Constll inf=1e13+7;structedge{int  from, to; intCost ;}; Edge ES[MAXM];p riority_queue<P,vector<P>,greater<P> >Que;vector<int>G[MAXN],T[MAXN];intPre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;stack<int>s;voidDfsintu) {Pre[u]=lowlink[u]=++Dfs_clock;    S.push (U);  for(intI=0; I<g[u].size (); i++)    {        intv=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++;  while(true)        {            intx=S.top ();            S.pop (); SCCNO[X]=scc_cnt; if(X==u) Break; }    }}voidFIND_SCC (intN) {Dfs_clock=scc_cnt=0; memset (Sccno,0,sizeof(SCCNO)); memset (PRE,0,sizeof(pre));  for(intI=1; i<=n; i++)        if(!Pre[i]) DFS (i);}voidBuildintm) {     for(intI=1; i<=scc_cnt; i++) t[i].clear ();  for(intI=1; i<=m; i++)    {        intU=es[i]. from, v=es[i].to; if(Sccno[u]==sccno[v])Continue;    T[sccno[u]].push_back (Sccno[v]); }}intCY[MAXN],VIS[MAXN];BOOLDFS2 (intu) {     for(intI=0; I<t[u].size (); i++)    {        intv=T[u][i]; if(Vis[v])Continue; VIS[V]=true; if(cy[v]==-1||DFS2 (Cy[v])) {Cy[v]=u; return true; }    }    return false;}intSolveintN) {    intret=0; memset (CY,-1,sizeof(CY));  for(intI=1; i<=n; i++) {memset (Vis,0,sizeof(VIS)); RET+=DFS2 (i); }    returnN-ret;}intMain () {intT; scanf ("%d",&t);  while(t--)    {        intn,m; scanf ("%d%d",&n,&m);  for(intI=1; i<=n;i++) g[i].clear ();  for(intI=1; i<=m; i++)        {            intu,v; scanf ("%d%d",&u,&v); Es[i]. from=u,es[i].to=v;        G[u].push_back (v);        } FIND_SCC (n);        Build (m); cout<<solve (scc_cnt) <<Endl; }    return 0;}

Tarjan indent + Minimum path overlay

HDU 3861.The King's Problem strong Unicom component + minimum path coverage