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;
- }
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