标签:hdu 二分图染色 贪心
wyh2000 and pupil
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1040 Accepted Submission(s): 331
Problem Description
Young theoretical computer scientist wyh2000 is teaching his pupils.
Wyh2000 has n pupils.Id of them are from 1
to n.In
order to increase the cohesion between pupils,wyh2000 decide to divide them into 2 groups.Each group has at least 1 pupil.
Now that some pupils don‘t know each other(if a
doesn‘t know b,then
b
doesn‘t know a).Wyh2000
hopes that if two pupils are in the same group,then they know each other,and the pupils of the first group must be as much as possible.
Please help wyh2000 determine the pupils of first group and second group. If there is no solution, print "Poor wyh".
Input
In the first line, there is an integer
T
indicates the number of test cases.
For each case, the first line contains two integers
n,m
indicate the number of pupil and the number of pupils don‘t konw each other.
In the next m lines,each line contains 2 intergers
x,y(x<y),indicates
that x
don‘t know y
and y
don‘t know x,the
pair (x,y)
will only appear once.
T≤10,0≤n,m≤100000
Output
For each case, output the answer.
Sample Input
2
8 5
3 4
5 6
1 2
5 8
3 5
5 4
2 3
4 5
3 4
2 4
Sample Output
Source
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5285
题目大意:一些人要分成两组,(这群人真的好累啊,一天到晚被分来分去),给出他们之间的不认识关系,两个不认识的人不能分到一组,且每组都要至少有一个人,问分完后两组人数的最大可能值和最小可能值
题目分析:题意化简一下,就是求几个独立的二分图(可能有单点不过不碍事)合成一个大二分图,使得大二分图两侧点数差的绝对值最大,显然这里用贪心思想,对每个二分图,我把点数大的那一侧的全部并起来作为大二分图的一侧,另一侧用n减即可,判断二分图和记录点数采用的是DFS染色法,注意几个wa点,n<2时肯定没法分,还有m等于0时,显然可以有n个平凡图,这种特殊情况下,因为每组至少有一个人,直接n-1和1即可
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 100005;
int n, m, cnt;
int head[MAX], color[MAX], num[2];
bool vis[MAX], flag;
struct EDGE
{
int v, next;
}e[MAX * 2];
void Add(int u, int v)
{
e[cnt].v = v;
e[cnt].next = head[u];
head[u] = cnt ++;
}
void DFS(int u, int col)
{
vis[u] = true;
color[u] = col;
num[col] ++;
for(int i = head[u]; i != -1; i = e[i].next)
{
int v = e[i].v;
if(vis[v])
{
if(color[v] == color[u])
{
flag = true;
return;
}
}
else
DFS(v, col ^ 1);
}
return;
}
int main()
{
int T;
scanf("%d", &T);
while(T --)
{
memset(head, -1, sizeof(head));
memset(color, -1, sizeof(color));
memset(vis, false, sizeof(vis));
cnt = 0;
flag = false;
scanf("%d %d", &n, &m);
for(int i = 0; i < m; i++)
{
int u, v;
scanf("%d %d", &u, &v);
Add(u, v);
Add(v, u);
}
if(n < 2)
{
printf("Poor wyh\n");
continue;
}
if(m == 0)
{
printf("%d 1\n", n - 1);
continue;
}
int ans = 0;
for(int i = 1; i <= n; i++)
{
if(!vis[i])
{
memset(num, 0, sizeof(num));
DFS(i, 0);
if(flag)
break;
ans += max(num[0], num[1]);
}
}
if(flag)
printf("Poor wyh\n");
else
printf("%d %d\n", ans, n - ans);
}
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
HDU 5285 wyh2000 and pupil (DFS染色判二分图 + 贪心)
标签:hdu 二分图染色 贪心
原文地址:http://blog.csdn.net/tc_to_top/article/details/47066435