# 畅通工程，How Many Tables ACM第九天-图论

下面两题的方法是一样的都是并查集的应用；
并查集的重点就是找到祖先节点的过程。
重点在这里：int find(int x){ return p[x] == x ? x : p[x] = find( p[x]); }
不断地递归，递归到找到自己是自己的祖先为止。并且把它取接到那个祖先节点上；这样可以节省空间和时间。

# How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15414    Accepted Submission(s): 7560

Problem Description
Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

Input
The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

Output
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

Sample Input

2 5 3 1 2 2 3 4 5 5 1 2 5

Sample Output

2 4

Author
Ignatius.L

Source

Recommend
Eddy

#include <cstdio>
#define maxn 1005
using namespace std;
int p[maxn];
int a[maxn];
int b[maxn];
int find(int x){
return	p[x] == x? x : p[x] = find(p[x]);
}
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,m;
scanf("%d %d",&n,&m);
for(int i = 1;i <= m;i++){
scanf("%d %d",&a[i],&b[i]);
}
int ans = 0;
for(int i = 1;i <= n;i++){
p[i] = i;
}
for(int i = 0;i <= m;i++){
int x = find(a[i]);
int y = find(b[i]);
if(x != y){
ans++;
p[x] = y;
}
}
int answer = n - ans;
}
return	0;
}

# 畅通工程

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32971    Accepted Submission(s): 17391

Problem Description

Input

3 3
1 2
1 2
2 1

Output

Sample Input

4 2 1 3 4 3 3 3 1 2 1 3 2 3 5 2 1 2 3 5 999 0 0

Sample Output

1 0 2 998
Hint
Hint
Huge input, scanf is recommended.

Source

Recommend
JGShining




#include <cstdio>
#include <cstring>

using namespace std;
#define maxn 1005
int p[maxn];
int a[maxn];
int b[maxn];
int find(int x){
return	p[x] == x? x : p[x] = find(p[x]);
}
int main(){
int n,m;
while(scanf("%d %d",&n,&m) && n){
memset(p,0,sizeof(p));
for(int i = 1;i <= m;i++){
scanf("%d %d",&a[i],&b[i]);
}
int ans = 0;
for(int i = 1;i <= n;i++){
p[i] = i;
}
for(int i = 0;i <= m;i++){
int x = find(a[i]);
int y = find(b[i]);
if(x != y){
ans++;
p[x] = y;
}
}
int answer = n - 1 - ans;
}
return	0;
}

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