Hdoj Topic 3861 The King's problem (strong connectivity, minimum point coverage)

The King ' s problemTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 1830 Accepted Submission (s): 666


Problem DescriptionIn The 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 Input

13 21 21 3

Sample Output

2

Source2011 multi-university Training Contest 3-host by BIT
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#include <stdio.h> #include <string.h> #include <vector> #include <string> #include < Iostream> #define MAX (A, B) (a>b?a:b) #define MIN (A, b) (a>b?b:a) using namespace Std;int dfn[5050],low[5050],ins [5050],head[5050],belong[5050],stack[5050],link[5050];int cnt,top,taj,time,n,m;vector<int>vt[5050];struct S{int U,v,next;} Edge[100010*2];void init () {Cnt=top=taj=time=0;memset (belong,-1,sizeof (belong)); memset (ins,0,sizeof (INS)); memset (Low,-1,sizeof (Low)); Memset (Dfn,-1,sizeof (DFN)); Memset (Head,-1,sizeof (head)); memset (stack,0,sizeof (stack));} void Add (int u,int v) {edge[cnt].u=u;edge[cnt].v=v;edge[cnt].next=head[u];head[u]=cnt++;} void Tarjin (int u) {dfn[u]=low[u]=time++;stack[top++]=u;ins[u]=1;for (int i=head[u];i!=-1;i=edge[i].next) {int V=edge [I].v;if (Dfn[v]==-1) {Tarjin (v); Low[u]=min (Low[u],low[v]);} ElseIf (Ins[v]) low[u]=min (Dfn[v],low[u]);} if (Dfn[u]==low[u]) {Taj++;int now;do{now=stack[--top];belong[now]=taj;ins[now]=0;} while (U!=now);}} int dfs (int u) {for (int i=0;i<vt[U].size (); i++) {int v=vt[u][i];if (!ins[v]) {ins[v]=1;if (link[v]==-1| | DFS (Link[v])) {Link[v]=u;return 1;}}} return 0;} int main () {int t;scanf ("%d", &t), while (t--) {int. i;init (); scanf ("%d%d", &n,&m); for (i=1;i<=n;i++) Vt[i] . Clear (); while (m--) {int u,v;scanf ("%d%d", &u,&v); add (u,v);} for (i=1;i<=n;i++) {if (dfn[i]==-1) {Tarjin (i);}} for (i=0;i<cnt;i++) {int u=edge[i].u;int v=edge[i].v;if (Belong[u]!=belong[v]) {vt[belong[u]].push_back (Belong[v] );//printf ("%d%d%d%d\n", U,v,belong[u],belong[v]);}} memset (link), int ans=0;for (i=1;i<=taj;i++) {memset (ins,0,sizeof (INS)), if (Dfs (i)) ans++;} printf ("%d\n", Taj-ans);}}

Hdoj Topic 3861 The King's problem (strong connectivity, minimum point coverage)