D. Connected Components
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
We already know of the large corporation where Polycarpus works as a system administrator. The computer network there consists of n computers and m cables that connect some pairs of computers. In other words, the computer network can be represented as some non-directed graph with n nodes and m edges. Let's index the computers with integers from 1 to n, let's index the cables with integers from 1 to m.
Polycarpus was given an important task — check the reliability of his company's network. For that Polycarpus decided to carry out a series of k experiments on the computer network, where the i-th experiment goes as follows:
- Temporarily disconnect the cables with indexes from l**i to r**i, inclusive (the other cables remain connected).
- Count the number of connected components in the graph that is defining the computer network at that moment.
- Re-connect the disconnected cables with indexes from l**i to r**i (that is, restore the initial network).
Help Polycarpus carry out all experiments and for each print the number of connected components in the graph that defines the computer network through the given experiment. Isolated vertex should be counted as single component.
Input
The first line contains two space-separated integers n, m (2 ≤ n ≤ 500; 1 ≤ m ≤ 104) — the number of computers and the number of cables, correspondingly.
The following m lines contain the cables' description. The i-th line contains space-separated pair of integers x**i, y**i (1 ≤ x**i, y**i ≤ n; x**i ≠ y**i) — the numbers of the computers that are connected by the i-th cable. Note that a pair of computers can be connected by multiple cables.
The next line contains integer k (1 ≤ k ≤ 2·104) — the number of experiments. Next k lines contain the experiments' descriptions. The i-th line contains space-separated integers l**i, r**i (1 ≤ l**i ≤ r**i ≤ m) — the numbers of the cables that Polycarpus disconnects during the i-th experiment.
Output
Print k numbers, the i-th number represents the number of connected components of the graph that defines the computer network during the i-th experiment.
Examples
input
Copy
6 51 25 42 33 13 661 32 51 55 52 43 3
output
Copy
456342
题意：
给你一个含有n个点，m个边的无向图。
以及q个询问
每一个询问，给定一个l和r，代表在原本的图中，删除e[l]~e[r] 这些边，
求剩下的图中联通快的个数。
思路：
我们建立2*m个并查集，
前m个是从1到m个边依次加入时的图网络联通情况，用并查集数组a表示
后m个维护反过来，即第m个到第1个边以此加入时的图网络联通情况。用并查集数组b来表示
对于每一个询问：
我们将a[l-1]和b[r+1]两个并查集合并，即可求得图中联通快的个数。
时间复杂度为\(O(n*m)\)
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline void getInt(int* p);
const int maxn = 10010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n, m;
struct dsu
{
int fa[505];
void init()
{
repd(i, 1, n)
{
fa[i] = i;
}
}
int findpar(int x)
{
if (fa[x] == x)
{
return x;
} else {
return fa[x] = findpar(fa[x]);
}
}
void mg(int a, int b)
{
a = findpar(a);
b = findpar(b);
if (a != b)
{
fa[a] = b;
}
}
int getans()
{
int res = 0;
repd(i, 1, n)
{
if (fa[i] == i)
{
res++;
}
}
return res;
}
} a[maxn], b[maxn];
dsu t1, t2;
pii c[maxn];
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
while (~du2(n, m))
{
a[0].init();
b[m + 1].init();
t1.init();
repd(i, 1, m)
{
du2(c[i].fi, c[i].se);
t1.mg(c[i].fi, c[i].se);
a[i] = t1;
}
t1.init();
for (int i = m; i >= 1; --i)
{
t1.mg(c[i].fi, c[i].se);
b[i] = t1;
}
int q;
scanf("%d", &q);
int l, r;
while (q--)
{
du2(l, r);
t2 = a[l - 1];
repd(i, 1, n)
{
// chu(t2.findpar(i));
// chu(b[r + 1].findpar(i));
t2.mg(t2.findpar(i), b[r + 1].findpar(i));
}
printf("%d\n", t2.getans() );
}
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
京公网安备 11010502036488号