HDU 3749 Financial Crisis(点-双连通分量)

Because of the financial crisis, a large number of enterprises go bankrupt. In addition to this, other enterprises, which have trade relation with the bankrup enterprises, are also faced with closing down. Owing to the market collapse, profit decline and funding chain intense, the debt-ridden entrepreneurs 
have to turn to the enterprise with stably developing for help. 

Nowadays, there exist a complex net of financial trade relationship between enterprises. So if one of enterprises on the same financial chain is faced with bankrupt, leading to cashflow's obstruction, the other enterprises related with it will be influenced as well. At the moment, the foresight entrepreneurs are expected the safer cooperation between enterprises. In this sense, they want to know how many financial chains between some pairs of enterprises are independent with each other. The indepence is defined that if there exist two roads which are made up of enterprises(Vertexs) and their financial trade relations(Edge) has the common start point S and end point T, and expect S and T, none of other enterprise in two chains is included in these two roads at the same time. So that if one of enterpirse bankrupt in one of road, the other will not be obstructed.

Now there are N enterprises, and have M pair of financial trade relations between them, the relations are Mutual. They need to ask about Q pairs of enterprises' safety situations. When two enterprises have two or more independent financial chains, we say they are safe enough, you needn't provide exact answers.

InputThe Input consists of multiple test cases. The first line of each test case contains three integers, N ( 3 <= N <= 5000 ), M ( 0 <= M <= 10000 ) and Q ( 1 <= Q <= 1000 ). which are the number of enterprises, the number of the financial trade relations and the number of queries. 
The next M lines, each line contains two integers, u, v ( 0 <= u, v < N && u != v ), which means enterpirse u and enterprise v have trade relations, you can assume that the input will not has parallel edge. 
The next Q lines, each line contains two integers, u, v ( 0 <= u, v < N && u != v ), which means entrepreneurs will ask you the financial safety of enterpirse u and enterprise v. 
The last test case is followed by three zeros on a single line, which means the end of the input.OutputFor each case, output the test case number formated as sample output. Then for each query, output "zero" if there is no independent financial chains between those two enterprises, output "one" if there is only one such chains, or output "two or more".Sample Input

3 1 2
0 1
0 2
1 0
4 4 2
0 1
0 2
1 2
2 3
1 2
1 3
0 0 0

Sample Output

Case 1:
zero
one
Case 2:
two or more
one

题意:

给你一个(保证输入无重边,无自环)无向图,然后有下面Q条询问,每条询问为:问你u点与v点之间有几条(除了首尾两点外,其他点不重复)的路径.如果有0条或1条输出0或1,如果有2条以上,输出”two or more”.

分析:

首先如果u点与v点不连通,直接输出0即可.(用并查集实现)

然后如果u点与v点属于同一个点-双连通分量,输出two or more.(这里有特例,两点一边的点-双连通分量应该输出1)

自己写的代码已知wrong不知道为什么，哪位大佬看一看谢谢。。

 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<map>
 using namespace std;

 int n,m,q,TAK,col,tim,top,tot;
 int a[],color[],dfn[],low[],stack[],ins[];
 int cnt,head[],Next[],rea[];
 struct Node
 {
     int x,y;
 }b[];
 map<int,int>p;

 void add(int u,int v)
 {
     Next[++cnt]=head[u];
     head[u]=cnt;
     rea[cnt]=v;
 }
 void Tarjan(int u,int fa)
 {
     int num=;color[u]=col;
     dfn[u]=low[u]=++tim;
     stack[++top]=u,ins[u]=;
     for (int i=head[u];i!=-;i=Next[i])
     {
         int v=rea[i];
         if (v==fa) continue;
         if (!dfn[v])
         {
             Tarjan(v,u);num++;
             low[u]=min(low[u],low[v]);
             if (low[v]>=dfn[u])
             {
                 tot=;
                 int x=-;
                 while(x!=u)
                 {
                     x=stack[top--];
                     ins[x]=;
                     a[++tot]=x;
                 }
                 stack[++top]=u,ins[u]=;
                 if (tot==) continue;
                 for (int j=;j<tot;j++)
                     for (int k=j+;k<=tot;k++)
                     {
                         int x=a[j],y=a[k];
                         if (x>y) swap(x,y);
                         if (p[x*n+y]==-) p[x*n+y]=;
                     }
             }
         }
         else low[u]=min(low[u],dfn[v]);
     }
 }
 int main()
 {
     while(~scanf("%d%d%d",&n,&m,&q)&&(n+m+q))
     {
         printf("Case %d:\n",++TAK);
         cnt=top=tim=;p.clear();
         memset(dfn,,sizeof(dfn));
         memset(low,,sizeof(low));
         memset(color,,sizeof(color));
         memset(head,-,sizeof(head));
         for (int i=,x,y;i<=m;i++)
         {
             scanf("%d%d",&x,&y);
             x++,y++;
             add(x,y),add(y,x);
         }
         for (int i=;i<=q;i++)
         {
             scanf("%d%d",&b[i].x,&b[i].y);
             b[i].x++,b[i].y++;
             if (b[i].x>b[i].y) swap(b[i].x,b[i].y);
             p[b[i].x*n+b[i].y]=-;
         }
         for (int i=;i<=n;i++)
             if (!dfn[i])
             {
                 col=i;
                 Tarjan(i,-);
             }
         for (int i=;i<=q;i++)
             if (color[b[i].x]!=color[b[i].y]) printf("zero\n");
             else if (p[b[i].x*n+b[i].y]==-) printf("one\n");
             else printf("two or more\n");
     }
 }