critical links(UVA-796)求桥的数目

题目:UVA-796

题意:给一无向图,求图中桥的个数;

题中要求输出具体路径;

所以用了vector;hash是用来判断重边的。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
#include <bits/stdc++.h>
using namespace std;

const int maxn = 10010;
const int maxm = 100010;
struct Edge
{
    int to,next;
    bool cut;
}edge[maxm];

int head[maxn],tot;
int Low[maxn],DFN[maxn],Stack[maxn];
int Index,top;
bool Instack[maxn];
bool cut[maxn];
int add_block[maxn];
int bridge;
void addedge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].cut = false;
    head[u] = tot ++;
}

void Tarjan(int u,int pre)
{
    int v;
    Low[u] = DFN[u] = ++ Index;
    Stack[top ++] = u;
    Instack[u] = true;
    int son = 0;
    int pre_cnt = 0;
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        v = edge[i].to;
        if(v == pre && pre_cnt == 0){
            pre_cnt ++;
            continue;
        }
        if(!DFN[v])
        {
            son ++;
            Tarjan(v,u);
            if(Low[u] > Low[v])Low[u] = Low[v];
            if(Low[v] > DFN[u])
            {
                bridge ++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
            if(u != pre && Low[v] >= DFN[u])
            {
                cut[u] = true;
                add_block[u] ++;
            }
        }
        else if(Low[u] > DFN[v])
            Low[u] = DFN[v];
    }
    if(u == pre && son > 1) cut[u] = true;
    if(u == pre) add_block[u] = son - 1;
    Instack[u] = false;
    top --;
}

void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(add_block,0,sizeof(add_block));
    memset(cut,false,sizeof(cut));
    Index = top = 0;
    bridge = 0;
    for(int i = 1;i <= N;i ++)
        if(!DFN[i])
            Tarjan(i,i);
    printf("%d critical links\n",bridge);
    vector<pair<int,int> >ans;
    for(int u = 1;u <= N;u ++)
    {
        for(int i = head[u];i != -1;i = edge[i].next)
            if(edge[i].cut && edge[i].to > u)
        {
            ans.push_back(make_pair(u,edge[i].to));
        }
    }
    sort(ans.begin(),ans.end());
    for(int i = 0;i < ans.size();i ++)
    {
        printf("%d - %d\n",ans[i].first - 1,ans[i].second - 1);
    }
    printf("\n");
}

void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
map<int,int>mapit;
inline bool isHash(int u ,int v)
{
    if(mapit[u * maxn + v]) return true;
    if(mapit[v * maxn + u]) return true;
    mapit[u * maxn + v] = mapit[v * maxn + u] = 1;
    return false;
}
int main()
{
    int n;
    while(cin >> n)
    {
        init();
        int u;
        int k;
        int v;
        for(int i = 1;i <= n;i ++)
        {
            scanf("%d (%d)",&u,&k);
            u ++;
            while(k --)
            {
                scanf("%d",&v);
                v ++;
                if(v <= u) continue;
                addedge(u,v);
                addedge(v,u);
            }
        }
        solve(n);
    }
    return 0;
}

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转载自blog.csdn.net/guagua_de_xiaohai/article/details/84713476