ZOJ 2588 Burning Bridges

 

大意：求一个无向图的割边，其中可能有重边。

思路：Tarjan求保存割边，排序输出。

不知咋的，数组开到题目给定的100010会报错，改成200020就行了，KD啊。

CODE：

 

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

#define MAXN 10010
#define MAXM 100010

struct Edge
{
    int v, next;
    int id;
}edge[MAXM*2];

int first[MAXN], low[MAXN], dfn[MAXN];
int bridge[MAXM*2];

int n, m;
int cnt;
int tot, nbridge;

void init()
{
    cnt = 0;
    tot = nbridge = 0;
    memset(first, -1, sizeof(first));
    memset(dfn, 0, sizeof(dfn));
}

void read_graph(int u, int v, int id)
{
    edge[cnt].v = v, edge[cnt].id = id;
    edge[cnt].next = first[u], first[u] = cnt++;
}

int check(int u, int fa)
{
    int count = 0;
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        int v = edge[e].v;
        if(v == fa) count++;
    }
    if(count >= 2) return 1;
    return 0;
}

void read_graph2()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        read_graph(u, v, i);
        read_graph(v, u, i);
    }
}

void Tarjan(int u, int fa)
{
    dfn[u] = low[u] = ++tot;
    bool repeat = 0;
    for(int e = first[u]; e !=- 1; e = edge[e].next)
    {
        int v = edge[e].v;
        if(v == fa && !repeat) //去掉一条反向边，其余的可能是重边。 
        {
            repeat = 1;
            continue;
        }
        if(!dfn[v])
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(dfn[u] < low[v])    bridge[nbridge++] = edge[e].id;
        }   //if(dfn[u] <= low[v]) 求割顶
        low[u] = min(low[u], dfn[v]);
    }
}

void solve()
{
    Tarjan(1, -1);
    sort(bridge, bridge+nbridge);
    printf("%d\n", nbridge);
    for(int i = 0; i < nbridge; i++)    printf(i != nbridge-1? "%d ":"%d\n", bridge[i]);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        init();
        read_graph2();
        solve();
        if(T) printf("\n"); 
    }
    return 0;
}

 

判重方法2：

 

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

#define MAXN 10010
#define MAXM 100010

struct Edge
{
    int v, next;
    int id;
}edge[MAXM*2];

int first[MAXN], low[MAXN], dfn[MAXN];
int bridge[MAXM*2];

int n, m;
int cnt;
int tot, nbridge;

void init()
{
    cnt = 0;
    tot = nbridge = 0;
    memset(first, -1, sizeof(first));
    memset(dfn, 0, sizeof(dfn));
}

void read_graph(int u, int v, int id)
{
    edge[cnt].v = v, edge[cnt].id = id;
    edge[cnt].next = first[u], first[u] = cnt++;
}

int check(int u, int fa)
{
    int count = 0;
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        int v = edge[e].v;
        if(v == fa) count++;
    }
    if(count >= 2) return 1;
    return 0;
}

void read_graph2()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        read_graph(u, v, i);
        read_graph(v, u, i);
    }
}

void Tarjan(int u, int fa)
{
    dfn[u] = low[u] = ++tot;
    bool repeat = 0;
    for(int e = first[u]; e !=- 1; e = edge[e].next)
    {
        int v = edge[e].v;
        if(v == fa && !repeat) //去掉一条反向边，其余的可能是重边。 
        {
            repeat = 1;
            continue;
        }
        if(!dfn[v])
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u])    bridge[nbridge++] = edge[e].id;
        }
        low[u] = min(low[u], dfn[v]);
    }
}

void solve()
{
    Tarjan(1, -1);
    sort(bridge, bridge+nbridge);
    printf("%d\n", nbridge);
    for(int i = 0; i < nbridge; i++)    printf(i != nbridge-1? "%d ":"%d\n", bridge[i]);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        init();
        read_graph2();
        solve();
        if(T) printf("\n"); 
    }
    return 0;
}