POJ 2762 infers single unicom (support points and even deflation + topological sequencing)

Going from U-V or from V to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14789 Accepted: 3915 Description In order to make their sons brave, Jiajia and wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, wind choose the rooms x and Y, and ask one of the their little sons go from one to the other. The son can either go from X-to-y, or from Y to X. Wind promised that her tasks is all possible, but she actually doesn ' t Know how to decide if a task is possible. To make hers life easier, Jiajia decided to choose a cave in which every pair of rooms are a possible task. Given a cave, can you tell Jiajia whether wind can randomly choose both rooms without worrying about anything? Input The first line contains a single integer T, the number of test cases. and followed T cases. The first line is contains integers n, m (0 < N < 1001,m < 6000), the number of rooms and corridors In the cave. The next m lines each contains to integers u and V, indicating that there is a corridor connecting the class U and the Class V dire ctly. Output The output should contain T lines. Write ' Yes ' if the cave have the property stated above, or ' No ' otherwise. Sample Input 13 31 22) 33 1 Sample Output Yes Source POJ Monthly--2006.02.26,zgl & TWB

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Test instructions: For random two node u,v. Assuming it can be from u to V or V to u, then output yes, otherwise output No.

train of thought: first strong connection contraction point, at this time must not contain the ring, see can find a longest road including all the contraction point can (actually find inference single chain), with topological sort OK, just assume there is only a strong connected component then must be yes, otherwise topo sort inference, code such as the following :

#include <iostream> #include <algorithm> #include <cstdio> #include <queue> #include < Cstring> #include <cmath> using namespace std;typedef long long ll;const int MAXN =1e4+10;const ll mod=20140413; Vector<int>g1[maxn],g2[maxn],g[maxn];int Sccno[maxn],vis[maxn],scc_cnt;//sccno Strong connected number, SCC_CNT represents the number of strongly connected components int Ind[maxn];//ind represents the in degree void init_g (int n)//Initialize the indent graph {memset (ind,0,sizeof ind); for (int i=1;i<=n;i++) g[sccno[i]].clear () ; for (int i=1;i<=n;i++) {for (int j=0;j<g1[i].size (); ++j) {int &v =g1[i][j];if (Sccno[i]!=sccno[v]) {g[sccno[i ]].push_back (Sccno[v]); ind[sccno[v]]++;}}} BOOL Toposort (int n)//topo sort {init_g (n); int u,cnt=0;queue<int>q;for (int i=1;i<=n;i++) {if (!ind[sccno[i]]) { if (!q.empty ()) return False;q.push (Sccno[i]); cnt++;}} while (!q.empty ()) {U=q.front (); Q.pop (); ind[u]=-1;for (int i=0;i<g[u].size (); i++) {int &v=g[u][i];ind[v]--;if (ind[v]==0) {if (!q.empty ()) return False;q.push (v); cnt++;}}} return cnt==scc_cnt;} Vector<int>s;void INIT (int n) {memset (vis,0,sizeof vis); Memset (sccno,0,sizeof sccno); S.clear (); scc_cnt=0;for (int i=1;i<=n;i++) {g1[i].clear (); G2[i].clear (); G[i].clear ();}} void Addedge (int u,int v) {g1[u].push_back (v); G2[v].push_back (u);} void dfs1 (int u) {if (Vis[u]) return, vis[u]=1;for (int i=0;i<g1[u].size (); i++) DFS1 (G1[u][i]); S.push_back (u);} void dfs2 (int u) {if (Sccno[u]) return, sccno[u]=scc_cnt;for (int i=0;i<g2[u].size (); ++i) DFS2 (G2[u][i]);} BOOL Is_semiconnect (int n)///calculates the strongly connected component, tentatively infers {for (int i=1;i<=n;i++) DFS1 (i), the for (int i=s.size (); i>=1;i--) {if (! Sccno[s[i-1]] {scc_cnt++;d fs2 (s[i-1]);}} return scc_cnt<=1;}    int main () {int t,n,m;   scanf ("%d", &t);   while (t--) {scanf ("%d%d", &n,&m);   int u,v;   Init (n);   while (m--) {scanf ("%d%d", &u,&v);   Addedge (U,V);   } if (Is_semiconnect (n)) puts ("Yes");   else{if (Toposort (n)) puts ("Yes");   Else puts ("No");  }} return 0; }

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POJ 2762 infers single unicom (support points and even deflation + topological sequencing)