用到的算法
割邊 + 縮點(得到邊連通分量) + 樸素LCA
算法解析
-
無向圖區分重邊與同一條邊的反方向: 對每一條邊都用一個變量id來標識,一條無向邊的兩個方向用同一個id表示。
-
割邊:
c++ if(low[v] > dfn[u])
,即以點v爲根的子樹不能到達點u及以上,所以邊uv爲一條割邊。 -
縮點(得到邊雙連通分量): 去掉橋後,圖就變成了若干個隔離的邊雙連通分量,因爲將這些分量縮點。
- 實現方法:直接對整個圖進行dfs,如果有些點屬於同一個連通分量,則可以被相同的變量值標記,代表同一個點。這樣以每一個沒有被標記過的點爲起點進行dfs,則完成縮點。
-
樸素LCA:離線查詢,時間複雜度是兩點距離最近公共祖先的距離和,
- 思想:若兩點的深度不一樣,則深度大的結點向上跳;若兩點多的深度一樣但不是同一個結點,則兩點同時向上跳。
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其他注意的點
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求得割邊後,如何找到它同一條邊反方向的邊?利用疑惑操作的性質,若a是奇數,則 a ^ 1 = a - 1; 若a是偶數,則 a ^ 1 = a + 1.
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由於LCA是對樹的操作, 所以新圖必須要找到一個根節點,來確定其它結點的深度。可以讓1號結點作爲根節點,因爲縮點時,新形成點的序號是從1開始的。
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添加邊後,需要對途徑的邊進行標記,防止下次再減一遍,但LCA是對點的操作,如何標記邊?判斷邊的下端點,若點需要向上跳且該點還未標記過,則割邊數--。
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//poj 3694
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <map>
#include <iostream>
using namespace std;
typedef long long LL;
const double pi = acos(-1.0);
const double e = exp(1);
//const int MAXN =2e5+10;
const LL N = 1000000007;
struct edge
{
int id;
int flag;
int from;
int to;
int next;
} edge[200009], edge3[200009];
int head[200009];
int head3[200009]; //描述新圖中結點之間的關係
int dfn[100009];
int low[100009];
int cnt = 1;
int cnt3 = 0; //用於構建縮點後新圖的鏈式前向星
int New[100009]; //第i個雙聯通分量所含有的結點個數
int captain[100009]; //點i所在的邊雙連通分量
int father[100009];
int vis[100009];
int deep[100009];
int sum;
void tarjan(int u, int id) //需要考慮去重邊的問題
{
int i;
low[u] = dfn[u] = cnt++;
for (i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(id == edge[i].id) //判斷是不是同一條邊,若id相同則爲同一條
continue;
if (!dfn[v])
{
tarjan(v, edge[i].id);
low[u] = min(low[u], low[v]);
if (low[v] > dfn[u])
{
edge[i].flag = edge[i ^ 1].flag = 1; // 標記割邊,在尋找邊雙連通分量時忽略掉該邊
// printf("%d ---> %d\n", edge[i].from, edge[i].to);
}
}
else
{
low[u] = min(low[u], dfn[v]);
}
}
}
void seek_doubleEdge(int u, int cnt2) //縮點
{
int i;
dfn[u] = cnt2;
New[cnt2]++;
captain[u] = cnt2;
for (i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].flag) //連接新形成的點
{
if (captain[v]) //如果所連接的點還沒有縮點,則跳過。由於是條無向邊且當前端點已經縮點,當對無向邊的另一個端點縮點時可以建立一條邊。
{
int a = captain[u];
int b = captain[v];
edge3[cnt3].to = b;
edge3[cnt3].next = head3[a];
head3[a] = cnt3++;
edge3[cnt3].to = a;
edge3[cnt3].next = head3[b];
head3[b] = cnt3++;
// cout << u << " " << v << " " << a << " "<< b << " ** " << endl;
}
continue;
}
else
{
if (!dfn[v])
{
seek_doubleEdge(v, cnt2);
}
}
}
}
void init_lca(int u, int fa, int d)
{
int i;
deep[u] = d;
father[u] = fa;
// printf("%d %d ?? \n", u, deep[u]);
for (i = head3[u]; i != -1; i = edge3[i].next)
{
int v = edge3[i].to;
if (!deep[v])
{
//printf("%d %d %d (^_^)\n", v, u, d);
init_lca(v, u, d + 1);
}
}
}
int lca(int a, int b)
{
while (deep[a] < deep[b])
{
if (vis[b] == 0 && b != 1)
{
sum--;
vis[b] = 1;
}
b = father[b];
}
while (deep[a] > deep[b])
{
if (vis[a] == 0 && a != 1)
{
sum--;
vis[a] = 1;
}
a = father[a];
}
while (deep[a] == deep[b] && a != b)
{
if (vis[a] == 0 && a != 1)
{
sum--;
vis[a] = 1;
}
if (vis[b] == 0 && b != 1)
{
sum--;
vis[b] = 1;
}
a = father[a];
b = father[b];
}
return sum;
}
int main()
{
int n, m, i;
int cnt1, a, b, q, nn = 0;
while (scanf("%d%d", &n, &m) != EOF)
{
if (n == 0 && m == 0)
break;
nn++;
cnt1 = 0;
cnt = 1;
memset(head, -1, sizeof(head));
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(captain, 0, sizeof(captain));
while (m--)
{
scanf("%d%d", &a, &b);
edge[cnt1].id = cnt1; //[1]
edge[cnt1].flag = 0;
edge[cnt1].from = a;
edge[cnt1].to = b;
edge[cnt1].next = head[a];
head[a] = cnt1++;
edge[cnt1].id = cnt1 - 1; //[2],同[1]處註釋一起爲一條無向邊的來回方向都標記同一個id.
edge[cnt1].flag = 0;
edge[cnt1].from = b;
edge[cnt1].to = a;
edge[cnt1].next = head[b];
head[b] = cnt1++;
}
for (i = 1; i <= n; i++) //雙連通分量去割邊
{
if (!dfn[i])
{
tarjan(i, -1);
}
}
memset(dfn, 0, sizeof(dfn));
memset(head3, -1, sizeof(head3));
memset(vis, 0, sizeof(vis));
memset(deep, 0, sizeof(deep));
int cnt2 = 0;
for (int i = 1; i <= n; i++) //標記雙連通分量(縮點)
{
if (!dfn[i])
{
cnt2++;
seek_doubleEdge(i, cnt2);
}
}
/*
cout << " ** " << cnt2 << endl; // 邊雙連通分量的個數
for (int i = 1; i <= n; i++)
{
printf("* %d %d\n", i, captain[i]);
}
for (int i = 1; i <= cnt2; i++) //輸出縮點後的新圖
{
for (int j = head3[i]; j != -1; j = edge3[j].next)
{
cout << i << " " << edge3[j].to << endl;
}
}
*/
sum = cnt2 - 1;
init_lca(1, -1, 1);
/* for (int i = 1; i <= cnt2; i++)
{
cout << i << " " << deep[i] << " ?? " << endl;
}
*/
printf("Case %d:\n", nn);
scanf("%d", &q);
while (q--)
{
scanf("%d%d", &a, &b);
if (captain[a] != captain[b])
{
lca(captain[a], captain[b]);
}
printf("%d\n", sum);
}
printf("\n");
}
return 0;
}