poj3177 && poj3352 Edge Two-connected component pinch point

Redundant Paths

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12676		Accepted: 5368

Description

In order to get from one of the F (1 <= f <= 5,000) grazing fields (which is numbered 1..F) to another field, Bessi E and the rest of the herd is forced to cross near the Tree of rotten Apples. The cows is now tired of often being forced to take a particular path and want to build some new paths so that they would Always has a choice of at least, separate routes between any pair of fields. They currently has at least one route between each pair of fields and want to has at least. Of course, they can only travel on official Paths when they move from one field to another.

Given a description of the current set of R (F-1 <= R <=) paths This each connect exactly both different field s, determine the minimum number of new paths (each of the which connects exactly.) that must is built so that there is at least, separate routes between any pair of fields. Routes is considered separate if they use none of the same paths, even if they visit the same intermediate field along th e-To.

There might already be more than one paths between the same pair of fields, and your may also build a new path that connect s the same fields as some and other path.

Input

Line 1:two space-separated integers:f and R

Lines 2..r+1:each line contains, space-separated integers which is the fields at the endpoints of some path.

Output

Line 1: A single integer which is the number of the new paths this must be built.

Sample Input

7 71 22 33 42 54 55 65 7

Sample Output

2 Test instructions: give you a picture without a direction, judging at least how many sides to add, in order to make any 2 points between at least 2 independent (no common edge) of the road; train of thought: the point in the same double connected component can be equivalent to a point, the original image becomes a tree, then the problem is converted into a tree and how many edges can become a double connected graph. Answer = (the number of sides of the tree with a degree of 1 + 1)/2;

/** author:sweat123 * Created time:2016/6/21 20:07:00 * File Name:main.cpp*/#include<Set>#include<map>#include<queue>#include<stack>#include<cmath>#include<string>#include<vector>#include<cstdio>#include<time.h>#include<cstring>#include<iostream>#include<algorithm>#defineINF 1<<30#defineMOD 1000000007#definell Long Long#defineLson l,m,rt<<1#defineRson m+1,r,rt<<1|1#definePi ACOs (-1.0)using namespacestd;Const intMAXN =5010;structnode{intto ; intNext; }EDGE[MAXN<<2];intPre[maxn],vis[maxn],pa[maxn],dfn[maxn],low[maxn],n,m,ind;intPX[MAXN],PY[MAXN];intpcnt;voidAddintXinty) {edge[ind].to=y; Edge[ind].next=Pre[x]; PRE[X]= ind + +; }intFindintx) {    if(pa[x]! = x) Pa[x] =find (pa[x]); returnPa[x]; }voidDfsintRtintKintFA) {Dfn[rt]= Low[rt] =K;  for(inti = Pre[rt]; I! =-1; i =Edge[i].next) {        intt =edge[i].to; if(!dfn[t] && t! =FA) {DFS (t,k+1, RT); LOW[RT]=min (low[rt],low[t]); if(Low[t] >Dfn[rt]) {px[pcnt]= rt,py[pcnt++] =T; } Else {                intFX =find (t); intFY =find (RT); if(PA[FX]! =Pa[fy]) {PA[FX]=fy; }            }         } Else if(t! = FA) {//Bridge is differenet from pointLOW[RT] =min (low[rt],dfn[t]); }    }   }intD[MAXN],F[MAXN];intMain () { while(~SCANF ("%d%d",&n,&m)) {IND=0; Pcnt=0; memset (PRE,-1,sizeof(pre));  for(inti =1; I <= m; i++){            intx, y; scanf ("%d%d",&x,&y);        Add (x, y), add (y,x); }           for(inti =1; I <= N; i++) {Pa[i]=i; } memset (DFN,0,sizeof(DFN)); memset (Low,0,sizeof(low)); DFS (1,1,-1); //for (int i = 1; I <= n; i++) {//cout<<dfn[i]<< ' <<low[i]<<endl; //}        //cout<<endl;memset (D,0,sizeof(d)); Memset (F,-1,sizeof(f)); intPnum =0;  for(inti =1; I <= N; i++){            intFX =find (i); if(F[FX] = =-1) F[FX] = + +Pnum; F[i]=F[FX]; }         for(inti =0; i < pcnt; i++) {D[f[px[i] ]++,d[f[py[i]] + +; }        intAns =0;  for(inti =1; I <= pnum; i++){            if(D[i] = =1) ans + +; } printf ("%d\n", (ans +1) /2); }    return 0;}

poj3177 && poj3352 Edge Two-connected component pinch point