HDU 1213 How Many Tables
2016-06-24 00:20
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213
Total Submission(s): 23102 Accepted Submission(s): 11517
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.
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.
5 3
1 2
2 3
4 5
5 1
2 5
4
思路:裸的并查集,直接上代码。
附上AC代码:
#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
const int maxn = 1005;
int p[maxn];
int n, m;
vector<int> v;
int _find(int x){
return p[x]==x ? x : p[x]=_find(p[x]);
}
int main(){
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int T;
scanf("%d", &T);
while (T--){
scanf("%d%d", &n, &m);
for (int i=1; i<=n; ++i)
p[i] = i;
int a, b;
while (m--){
scanf("%d%d", &a, &b);
a = _find(a);
b = _find(b);
p[a] = b;
}
int ans = 1;
v.clear();
v.push_back(_find(1));
for (int i=2; i<=n; ++i){
int t = _find(i);
if (find(v.begin(), v.end(), t) == v.end()){
++ans;
v.push_back(t);
}
}
printf("%d\n", ans);
}
return 0;
}
How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23102 Accepted Submission(s): 11517
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, andall 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 offriends, 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
25 3
1 2
2 3
4 5
5 1
2 5
Sample Output
24
Author
Ignatius.LSource
杭电ACM省赛集训队选拔赛之热身赛Recommend
Eddy思路:裸的并查集,直接上代码。
附上AC代码:
#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
const int maxn = 1005;
int p[maxn];
int n, m;
vector<int> v;
int _find(int x){
return p[x]==x ? x : p[x]=_find(p[x]);
}
int main(){
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int T;
scanf("%d", &T);
while (T--){
scanf("%d%d", &n, &m);
for (int i=1; i<=n; ++i)
p[i] = i;
int a, b;
while (m--){
scanf("%d%d", &a, &b);
a = _find(a);
b = _find(b);
p[a] = b;
}
int ans = 1;
v.clear();
v.push_back(_find(1));
for (int i=2; i<=n; ++i){
int t = _find(i);
if (find(v.begin(), v.end(), t) == v.end()){
++ans;
v.push_back(t);
}
}
printf("%d\n", ans);
}
return 0;
}
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