POJ3189 Steady Cow Assignment
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6817 | Accepted: 2349 |
Description
FJ would like to rearrange the cows such that the cows are as equally happy as possible, even if that means all the cows hate their assigned barn.
Each cow gives FJ the order in which she prefers the barns. A cow's happiness with a particular assignment is her ranking of her barn. Your job is to find an assignment of cows to barns such that no barn's capacity is exceeded and the size of the range (i.e.,
one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen) of barn rankings the cows give their assigned barns is as small as possible.
Input
Lines 2..N+1: Each line contains B space-separated integers which are exactly 1..B sorted into some order. The first integer on line i+1 is the number of the cow i's top-choice barn, the second integer on that line is the number of the i'th cow's second-choice
barn, and so on.
Line N+2: B space-separated integers, respectively the capacity of the first barn, then the capacity of the second, and so on. The sum of these numbers is guaranteed to be at least N.
Output
Sample Input
6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2
Sample Output
2
Hint
Each cow can be assigned to her first or second choice: barn 1 gets cows 1 and 5, barn 2 gets cow 2, barn 3 gets cow 4, and barn 4 gets cows 3 and 6.
Source
题目的意思是:有n头奶牛,m个棚,每个奶牛对每个棚都有一个喜爱程度。棚子有最大容量了,现在要给每个奶牛安家,找一个奶牛喜爱程度差值最小的方案问喜爱程度的区间最小为多大?
思路:尺取枚举区间端点,二分图多重匹配验证
注意:2~N+1行每行的每个数x不是指i对j的喜爱程度为x而是i对x的喜爱程度为j
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std; #define LL long long
const int INF = 0x3f3f3f3f;
const int MAXN=1005;
int uN,vN; //u,v数目
int g[MAXN][MAXN];
int linker[MAXN][MAXN];
bool used[MAXN];
int linknum[MAXN];
int cap[MAXN];
int mp[MAXN][MAXN];
int L,R; bool dfs(int u)
{
int v;
for(v=1; v<=vN; v++)
if(mp[u][v]<=R&&mp[u][v]>=L&&!used[v])
{
used[v]=true;
if(linknum[v]<cap[v])
{
linker[v][++linknum[v]]=u;
return true;
}
for(int i=1; i<=cap[v]; i++)
if(dfs(linker[v][i]))
{
linker[v][i]=u;
return true;
}
}
return false;
} int hungary()
{
int res=0;
int u;
memset(linknum,0,sizeof linknum);
memset(linker,-1,sizeof linker);
for(u=1; u<=uN; u++)
{
memset(used,0,sizeof used);
if(dfs(u)) res++;
}
return res;
} int main()
{
int n,m,k,x;
while(~scanf("%d%d",&uN,&vN))
{
for(int i=1; i<=uN; i++)
for(int j=1; j<=vN; j++)
{
scanf("%d",&x);
mp[i][x]=j;
} for(int i=1; i<=vN; i++)
scanf("%d",&cap[i]);
L= R = 1;
int ans = INF;
while(L <= R && R <= vN)
{
if(hungary()==uN)
{
ans=min(ans,R-L+1);
L++;
}
else
R++;
}
printf("%d\n",ans);
}
return 0;
}
POJ3189 Steady Cow Assignment的更多相关文章
- POJ3189 Steady Cow Assignment —— 二分图多重匹配/最大流 + 二分
题目链接:https://vjudge.net/problem/POJ-3189 Steady Cow Assignment Time Limit: 1000MS Memory Limit: 65 ...
- POJ3189 Steady Cow Assignment(最大流)
题目大概说,有n头牛和b块草地,每头牛心中分别对每块草地都有排名,草地在牛中排名越高牛安排在那的幸福度就越小(...),每块草地都能容纳一定数量的牛.现在要给这n头牛分配草地,牛中的幸福度最大与幸福度 ...
- POJ 2289 Jamie's Contact Groups & POJ3189 Steady Cow Assignment
这两道题目都是多重二分匹配+枚举的做法,或者可以用网络流,实际上二分匹配也就实质是网络流,通过枚举区间,然后建立相应的图,判断该区间是否符合要求,并进一步缩小范围,直到求出解.不同之处在对是否满足条件 ...
- POJ3189:Steady Cow Assignment(二分+二分图多重匹配)
Steady Cow Assignment Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7482 Accepted: ...
- POJ 3189——Steady Cow Assignment——————【多重匹配、二分枚举区间长度】
Steady Cow Assignment Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I ...
- Poj 3189 Steady Cow Assignment (多重匹配)
题目链接: Poj 3189 Steady Cow Assignment 题目描述: 有n头奶牛,m个棚,每个奶牛对每个棚都有一个喜爱程度.当然啦,棚子也是有脾气的,并不是奶牛想住进来就住进来,超出棚 ...
- Steady Cow Assignment POJ - 3189 (最大流+匹配)
Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which ...
- P2857 [USACO06FEB]稳定奶牛分配Steady Cow Assignment
题目描述 Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns ...
- Steady Cow Assignment
poj3189:http://poj.org/problem?id=3189 题意:这一题的题意.我看了很长时间才弄懂.就是给你n头牛,m个牛棚,每个牛对每一个牛棚会有一个满值,第i行第j个数表示的是 ...
随机推荐
- javascript 高级程序设计 三
Sorry,前两张介绍的主题还是JavaScript,而第一章介绍了JavaScript和ECMAScript区别,所以前两章介绍的主题应该改为ECMAScript,但是 标题就不改了因为现在人们习惯 ...
- RavenDb使用
在Raven中查询数据,查询条件必须在index中. 如果查询条件不在index中就会出现如下异常 var query = session.DynamicIndexQuery<ServicePr ...
- 20172306《Java程序设计》第四周学习总结
20172306 <Java程序设计>第四周学习总结 教材学习内容总结 第四章: 1. 类和对象的回顾:除了看书,我还上网找了一下两者的一些区别. 2. 编写类时,了解到初始化.形式参数. ...
- 201621123008 《Java程序设计》第十周学习总结
1. 本周学习总结 1.1 以你喜欢的方式(思维导图或其他)归纳总结异常相关内容. 2. 书面作业 本次PTA作业题集异常 1. 常用异常 结合题集题目7-1回答 1.1 自己以前编写的代码中经常出现 ...
- [转]Firefox+Burpsuite抓包配置(可抓取https)
0x00 以前一直用的是火狐的autoproxy代理插件配合burpsuite抓包 但是最近经常碰到开了代理却抓不到包的情况 就换了Chrome的SwitchyOmega插件抓包 但是火狐不能抓包的问 ...
- 【附源文件】软件工具类Web原型制作分享 - Sketch
Sketch是一款轻量,易用的矢量设计工具.专门为UI设计师开发,让UI设计更简单.更高效. 本原型由国产原型工具-Mockplus制作完成. 非常适合工具类产品官网使用,本模板的交互有通过使用面板组 ...
- js 正则表达式,匹配邮箱/手机号/用户名
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...
- 很漂亮的IP头格式
IP数据包格式 TCP/IP协议定义了一个在因特网上传输的包,称为IP数据报(IP Datagram).这是一个与硬件无关的虚拟包,由首部和数据两部分组成.首部的前一部分是固定长度,共 20 字节,是 ...
- 尼克的任务(P1280)
题目链接:尼克的任务 这道题,有点难度,也不是太难,因为我都做出来了. 好,下面分析一下: 这道题,显然的动规,我们这样设计状态. 我们设d[i]为从第i分钟初开始到结束有多少空闲时间. 那么我们的转 ...
- ServiceDesk Plus更有序地组织IT项目