HDOJ 题目3749 Financial Crisis（双联通）

Financial Crisis

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 212    Accepted Submission(s): 65

Problem Description

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.

 

Input

The 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.

 

Output

For 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
  

 

Author

possessor WC

 

Source

HDU 3rd “Vegetable-Birds Cup” Programming Open Contest

 

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lcy   |   We have carefully selected several similar problems for you:   3744  3751  3750  3747  3752 

 

转：http://blog.youkuaiyun.com/sdj222555/article/details/13746461

给一个无向图, n <5000, m < 10000

然后给出若干的询问（< 1000），问一个点到另一个点之间有多少条路可以走，只需回答有0条还是1条还是多条。注意这些路之间不能有边相同。

方法：

求点的双连通分量。

然后一个割点，有可能属于多个点双连通分量。

所以我们要是用vector把每个点属于的点双连通分量的编号都存起来。

然后我们要计算每个点双连通分量中的边的个数。 因为有那种只有一条边的双连通分量。

计算的方法就是查看边得两个端点所属的双连通分量，如果两个端点有同属于的一个双连通分量，就把对应的数目加1

对于每个询问。

首先看两个点是否连通。用并查集判断即可

然后看是否同属于一个双连通分量。 不属于就只能有一条路

然后同属于一个双连通分量的话，就看边得个数是否大于1。即可

ac代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <sstream>
#include <queue>
#include <vector>
#define MAXN 111111
#define MAXM 211111
#define eps 1e-8
#define INF 1000000001
using namespace std;
int e, tmpdfn, top;
int n, m, ind;
int vis[MAXM], head[MAXN], dfn[MAXN], low[MAXN], num[MAXM];
vector<int>g[MAXN];
struct Edge
{
    int v, next;
}edge[MAXM];
void add(int u, int v)
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
    edge[e].v = u;
    edge[e].next = head[v];
    head[v] = e++;
}
struct Stack
{
    int s, e;
    Stack(){}
    Stack(int a, int b){s = a; e = b;}
}st[MAXM];
void init()
{
    e = tmpdfn = ind = 0;
    top = -1;
    memset(vis, 0, sizeof(vis));
    memset(dfn, 0, sizeof(dfn));
    memset(head, -1, sizeof(head));
    memset(num, 0, sizeof(num));
}
void color(Stack t)
{
    ind++;
    while(top >= 0)
    {
        Stack A = st[top--];
        g[A.s].push_back(ind);
        g[A.e].push_back(ind);
        if(A.s == t.s && A.e == t.e) break;
    }
}
void dfs(int u, int fa)
{
    dfn[u] = low[u] = ++tmpdfn;
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if(vis[i] == 0)
        {
            vis[i] = vis[i ^ 1] = 1;
            Stack tmp(u, v);
            st[++top] = tmp;
            if(!dfn[v])
            {
                dfs(v, u);
                low[u] = min(low[u], low[v]);
                if(low[v] >= dfn[u]) color(tmp);
            }
            else if(v != fa) low[u] = min(low[u], dfn[v]);
        }
    }
}
int Q;
int uu[MAXM], vv[MAXM];
int fa[MAXN];
int find(int x)
{
    if(fa[x] == x) return x;
    int t = find(fa[x]);
    fa[x] = t;
    return t;
}
void join(int u, int v)
{
    int fx = find(u);
    int fy = find(v);
    if(fx != fy) fa[fx] = fy;
}
int main()
{
    int u, v;
    int cas = 0;
    while(scanf("%d%d%d", &n, &m, &Q) != EOF)
    {
        if(n == 0 && m == 0 && Q == 0) break;
        init();
        for(int i = 1; i <= n; i++) fa[i] = i;
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d", &uu[i], &vv[i]);
            uu[i]++; vv[i]++;
            add(uu[i], vv[i]);
            join(uu[i], vv[i]);
        }
        for(int i = 1; i <= n; i++) g[i].clear();
        for(int i = 1; i <= n; i++)
        {
            if(!dfn[i]) dfs(i, 0);
        }
        for(int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end());
        for(int i = 1; i <= n; i++) unique(g[i].begin(), g[i].end());

        for(int i = 0; i < m; i++)
        {
            int sz1 = g[uu[i]].size();
            int sz2 = g[vv[i]].size();
            int k1 = 0, k2 = 0;
            while(k1 < sz1 && k2 < sz2)
            {
                if(g[uu[i]][k1] > g[vv[i]][k2]) k2++;
                else if(g[uu[i]][k1] < g[vv[i]][k2]) k1++;
                else if(g[uu[i]][k1] == g[vv[i]][k2])
                {
                    int tmp = g[uu[i]][k1];
                    num[tmp]++;
                    break;
                }
            }
        }
        printf("Case %d:\n", ++cas);
        for(int i = 0; i < Q; i++)
        {
            scanf("%d%d", &u, &v);
            u++; v++;
            int sz1 = g[u].size();
            int sz2 = g[v].size();
            int k1 = 0, k2 = 0;
            int tmp = -1;
            while(k1 < sz1 && k2 < sz2)
            {
                if(g[u][k1] > g[v][k2]) k2++;
                else if(g[u][k1] < g[v][k2]) k1++;
                else if(g[u][k1] == g[v][k2])
                {
                    tmp = g[u][k1];
                    break;
                }
            }
            if(find(u) != find(v)) puts("zero");
            else if(tmp == -1) puts("one");
            else if(num[tmp] == 1) puts("one");
            else puts("two or more");
        }
    }
    return 0;
}