题意:给定一个无向连通的公路网,当某些公路路段检修的时候可能会由于该段公路不通,可能会使某些旅游点之间无法通行,求最少需要新建多少条公路,使得任意对一段公路进行检修的时候,所有的旅游景点之间仍然畅通。
思路:构造双连通图,
具体步骤:
[构造双连通图]
一个有桥的连通图,如何把它通过加边变成边双连通图?方法为首先求出所有的桥,然后删除这些桥边,剩下的每个连通块都是一个双连通子图。把每个双连通子图收缩为一个顶点,再把桥边加回来,最后的这个图一定是一棵树,边连通度为1。
统计出树中度为1的节点的个数,即为叶节点的个数,记为leaf。则至少在树上添加(leaf+1)/2条边,就能使树达到边二连通,所以至少添加的边数就是(leaf+1)/2。具体方法为,首先把两个最近公共祖先最远的两个叶节点之间连接一条边,这样可以把这两个点到祖先的路径上所有点收缩到一起,因为一个形成的环一定是双连通的。然后再找两个最近公共祖先最远的两个叶节点,这样一对一对找完,恰好是(leaf+1)/2次,把所有点收缩到了一起。
详细请见:http://www.byvoid.com/blog/biconnect/
CODE:
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define MAXN 50010
#define MAXM 100010
struct Edge
{
int v, next;
}edge[MAXM];
int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM], ind[MAXN];
int n, m;
int cnt;
int scnt, top, tot;
int leaf;
void init()
{
cnt = 0;
scnt = top = tot = 0;
leaf = 0;
memset(first, -1, sizeof(first));
memset(dfn, 0, sizeof(dfn));
memset(ins, 0, sizeof(ins));
memset(ind, 0, sizeof(ind));
}
void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u], first[u] = cnt++;
}
void Tarjan(int u, int father)
{
int v;
low[u] = dfn[u] = ++tot;
ins[u] = 1;
stack[top++] = u;
for(int e = first[u]; e != -1; e = edge[e].next)
{
v = edge[e].v;
if(v == father) continue; //与Tarjan求强连通分量的区别
if(!dfn[v])
{
Tarjan(v, u);
low[u] = min(low[u], low[v]);
}
else if(ins[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
scnt++;
do
{
v = stack[--top];
belong[v] = scnt;
ins[v] = 0;
}while(v != u);
}
}
void solve()
{
init();
while(m--)
{
int u, v;
scanf("%d%d", &u, &v);
read_graph(u, v);
read_graph(v, u);
}
Tarjan(1, -1); //Tarjan
for(int u = 1; u <= n; u++)
{
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(belong[u] != belong[v])
{
ind[belong[u]]++;
ind[belong[v]]++;
}
}
}
for(int i = 1; i <= scnt; i++) if(ind[i] == 2) leaf++; //由于每个点重复计算了一次,所以叶子节点的度为2
printf("%d\n", (leaf+1)/2);
}
int main()
{
while(~scanf("%d%d", &n, &m))
{
solve();
}
return 0;
}
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define MAXN 50010
#define MAXM 100010
struct Edge
{
int v, next;
}edge[MAXM];
int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM], ind[MAXN];
int n, m;
int cnt;
int scnt, top, tot;
int leaf;
void init()
{
cnt = 0;
scnt = top = tot = 0;
leaf = 0;
memset(first, -1, sizeof(first));
memset(dfn, 0, sizeof(dfn));
memset(ins, 0, sizeof(ins));
memset(ind, 0, sizeof(ind));
}
void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u], first[u] = cnt++;
}
void Tarjan(int u, int father)
{
int v;
low[u] = dfn[u] = ++tot;
ins[u] = 1;
stack[top++] = u;
for(int e = first[u]; e != -1; e = edge[e].next)
{
v = edge[e].v;
if(v == father) continue; //与Tarjan求强连通分量的区别
if(!dfn[v])
{
Tarjan(v, u);
low[u] = min(low[u], low[v]);
}
else if(ins[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
scnt++;
do
{
v = stack[--top];
belong[v] = scnt;
ins[v] = 0;
}while(v != u);
}
}
void solve()
{
init();
while(m--)
{
int u, v;
scanf("%d%d", &u, &v);
read_graph(u, v);
read_graph(v, u);
}
Tarjan(1, -1); //Tarjan
for(int u = 1; u <= n; u++)
{
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(belong[u] != belong[v])
{
ind[belong[u]]++;
ind[belong[v]]++;
}
}
}
for(int i = 1; i <= scnt; i++) if(ind[i] == 2) leaf++; //由于每个点重复计算了一次,所以叶子节点的度为2
printf("%d\n", (leaf+1)/2);
}
int main()
{
while(~scanf("%d%d", &n, &m))
{
solve();
}
return 0;
}