hdu2460 poj3694 求橋 + 求lca
阿新 • • 發佈:2019-01-05
http://poj.org/problem?id=3694
題意:
給定一個圖,對這個圖進行加邊,求出每加一條邊後,圖中橋的個數。
思路:
首先肯定是要先求出原圖中橋的個數,並且把橋標記起來。思考加一條邊以後橋的數量會減少多少,聯想到之前的那道題,就是加一條邊,能夠使橋的數量最少是多少。之前那個做法就是縮點後樹的直徑,這個就是可以減少的最多的橋的數量。因為如果這是一條鏈,將兩個端點連起來,這上面的橋都消失了。。
所以按照這個思路,可以考慮,對於每個要加的邊,求出他們的lca,在這個路徑上如果碰到一個橋就減一。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <cmath>
using namespace std;
const int M_node = 100009;
int low[M_node],dfn[M_node],father[M_node];
bool isbridge[M_node]; //標記橋 樹邊<u,v> 是橋則 isbridge[v] = true
int dfs_clock,num_br;
int n,m;
vector<int> G[M_node];
void init()
{
for(int i = 0; i <= n;i++) G[i].clear();
memset(dfn,0,sizeof(dfn));
memset(isbridge,false,sizeof(isbridge));
num_br = dfs_clock = 0;
}
int dfs(int u,int fa)
{
father[u] = fa;
int lowu = dfn[u] = ++dfs_clock;
for(int i = 0;i < G[u].size();i++)
{
int v = G[u][i];
if(!dfn[v])
{
int lowv = dfs(v,u);
lowu = min(lowu,lowv);
if(lowv > dfn[u])
{
num_br++;
isbridge[v] = true;
}
}
else if(dfn[v] < dfn[u] && v != fa) lowu = min(lowu,dfn[v]);
}
low[u] = lowu;
return lowu;
}
void lca(int u,int v)
{
while(dfn[u] > dfn[v])
{
if(isbridge[u])
{
num_br--;
isbridge[u] = false;
}
u = father[u];
}
while(dfn[v] > dfn[u])
{
if(isbridge[v])
{
num_br--;
isbridge[v] = false;
}
v = father[v];
}
while(u != v)
{
if(isbridge[u])
{
num_br--;
isbridge[u] = false;
}
if(isbridge[v])
{
num_br--;
isbridge[v] = false;
}
u = father[u];
v = father[v];
}
}
int main()
{
int k = 1;
//freopen("out.txt","w",stdout);
while(scanf("%d %d",&n,&m) == 2)
{
if(n == 0 || m == 0) break;
init();
for(int i = 0;i < m;i++)
{
int a,b;
scanf("%d %d",&a,&b);
G[a].push_back(b);
G[b].push_back(a);
}
dfs(1,-1);
int q;
scanf("%d",&q);
printf("Case %d:\n",k++);
while(q--)
{
int a,b;
scanf("%d %d",&a,&b);
lca(a,b);
printf("%d\n",num_br);
}
printf("\n");
}
return 0;
}