Codeforces Round #286 (Div. 1 B)

Http://codeforces.com/contest/506/problem/BProblem a graph also has n points, to M-relations < A, B, to indicate that a must arrive at B., the relationship satisfies transitivity, i.e. a->b,b->c , then A->c. At the same time, a to b,b not necessarily to a. Ask, at least how many edges to add, so that each relationship is set up. Solution for an n-point graph, because the edge is transitive, it may be possible to connect the n points into a ring, then must be 22 to each other, but this is not necessarily the least. For a weak link graph with n points, if it does not have a ring (it is a dag), then only the n-1 edge is needed (think about whether n points can be turned into a "chain" after you sort the DAG topology); if it exists, then turn it into a ring and need n edges. The algorithm is as follows: According to the given m relationships, the a->b constructs an edge. DFS, judging whether each weak link graph has a ring, then the statistical answer. In which to use, DFS awarded to the graph ring, and check the integration of weak unicom map. My Code

//hello. I ' m Peter.#pragma COMMENT (linker, "/stack:102400000,102400000")#include <cstdio>#include <iostream>#include <sstream>#include <cstring>#include <string>#include <cmath>#include <cstdlib>#include <algorithm>#include <functional>#include <cctype>#include <ctime>#include <stack>#include <queue>#include <vector>#include <set>#include <map>using namespace STD;typedef Long Longll#define Peter cout<< "I am Peter" <<endl#define INPUT Freopen ("Data.txt", "R", stdin)#define Randin srand ((unsigned int) time (NULL))#define INT (0x3f3f3f3f) * *#define LL (0x3f3f3f3f3f3f3f3f)#define MAXN 100010#define N 100100#define Minline intRead () {intx=0, f=1;CharCh=getchar (); while(ch>' 9 '|| ch<' 0 '){if(ch=='-') f=-1; Ch=getchar ();} while(ch>=' 0 '&&ch<=' 9 ') {x=x*Ten+ch-' 0 '; Ch=getchar ();}returnX*f;}intN,m;intNex[n],nexnum[n];inline voidInit_nex () { for(intI=1; i<=n;i++) {nex[i]=i; nexnum[i]=1; }}inline intFindnex (intx) {if(nex[x]!=x) Nex[x]=findnex (nex[x]);returnNEX[X];}structedge{intFrom,to,next;} edge[maxn<<1];intHead[n],num_edge;inline voidInit_edge () { for(intI=1; i<=n;i++) {head[i]=-1; } num_edge=0;}inline voidAdd_edge (intFromintTo) {intT=++num_edge;    Edge[t].from=from;    Edge[t].to=to;    Edge[t].next=head[from]; head[from]=t;}intVis[n],loop[n];BOOLOkinline voidDfsintNow) {if(!ok)return; vis[now]=1; for(inti=head[now];i!=-1; i=edge[i].next) {if(!ok)return;intto=edge[i].to;if(!vis[to]) DFS (to);Else if(vis[to]==1) {ok=false;return; }Else if(vis[to]==-1)Continue; } vis[now]=-1;}intMain () {N=read (), M=read ();    Init_nex (); Init_edge (); for(intI=1; i<=m;i++) {intA=read (), B=read (); Add_edge (A, b);intNex1=findnex (a);intNex2=findnex (b);if(NEX1!=NEX2)            {nex[nex1]=nex2;        NEXNUM[NEX2]+=NEXNUM[NEX1]; }    } for(intI=1; i<=n;i++) {intNex=findnex (i);if(loop[nex]| | Vis[i])Continue; ok=true; DFS (i);if(!ok) loop[nex]=true; }intans=0; for(intI=1; i<=n;i++) {intNex=findnex (i);if(vis[nex]!=-2) {vis[nex]=-2;if(Loop[nex]) Ans+=nexnum[nex];Elseans+=nexnum[nex]-1; }    }printf("%d\n", ans);return 0;}

Codeforces Round #286 (Div. 1 B)