POJ3189 Steady Cow Assignment
来源:互联网 发布:女神联盟2进阶15数据 编辑:程序博客网 时间:2024/05/29 07:44
Steady Cow Assignment
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6817 Accepted: 2349
Description
Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows really like their current barn, and some are not so happy.
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.
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
Line 1: Two space-separated integers, N and B
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.
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
Line 1: One integer, the size of the minumum range of barn rankings the cows give their assigned barns, including the endpoints.
Sample Input
6 41 2 3 42 3 1 44 2 3 13 1 2 41 3 4 21 4 2 32 1 3 2
Sample Output
2
Hint
Explanation of the sample:
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.
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
USACO 2006 February Gold
————————————————————————————
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 longconst 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;}
阅读全文
0 0
- POJ3189-Steady Cow Assignment
- POJ3189 Steady Cow Assignment
- POJ3189 Steady Cow Assignment
- 解题报告 之 POJ3189 Steady Cow Assignment
- POJ3189 Steady Cow Assignment 二分枚举+dinic最大流
- poj3189 Steady Cow Assignment --- 多重匹配,二分图匹配解法
- poj3189 Steady Cow Assignment(枚举+多重匹配)
- poj3189--Steady Cow Assignment (二分多重匹配)
- POJ 2289 Jamie's Contact Groups & POJ3189 Steady Cow Assignment
- Steady Cow Assignment
- 3189 Steady Cow Assignment //MaxMatch
- POJ 3189 Steady Cow Assignment
- POJ 3189 Steady Cow Assignment
- Steady Cow Assignment POJ - 3189
- POJ 3189 Steady Cow Assignment【网络流】
- POJ 3189 Steady Cow Assignment 笔记
- poj 3189 Steady Cow Assignment 枚举+网络流
- POJ 3189 Steady Cow Assignment 二分最大流
- <五>RecycleView+CardView实现瀑布流(类in界面效果)
- 在同一form表单中如何提交两个不同的action
- MyEclipse修改jsp模板的字符集
- MVC4.0网站发布和部署到IIS7.0上的方法
- ubuntu下composer的安装
- POJ3189 Steady Cow Assignment
- iOS 提示Revoke certificate 解决方案
- Android中跨进程通讯的4种方式
- 课程设计哈夫曼编/译码系统
- 浅析javaScript中的浅拷贝和深拷贝
- dubbo 远程服务调用流程
- 网络端口的分类
- Flask: 链接、静态文件
- TCP协议 标记位&定时器&三次握手四次挥手