zoukankan      html  css  js  c++  java
  • POJ3189 Steady Cow Assignment

    Steady Cow Assignment

    Language:
    Steady Cow Assignment
    Time Limit: 1000MSMemory Limit: 65536K
    Total Submissions: 7593Accepted: 2616

    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.

    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.

    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 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

    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.

    Source

    N头牛M个棚,每头牛对每个棚都有喜爱顺序,每个棚都有自己的容量
    问你怎么分配使得所有牛的喜爱程度差距最小

    题解

    二分答案,扫描建边,跑最大流来做多重匹配判断可行性。

    时间复杂度(O(sqrt{n+m}nm*mlog m))

    #include<iostream>
    #include<cstring>
    #include<queue>
    #define rg register
    #define il inline
    #define co const
    template<class T>il T read(){
        rg T data=0,w=1;rg char ch=getchar();
        for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
        for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
        return data*w;
    }
    template<class T>il T read(rg T&x) {return x=read<T>();}
    typedef long long ll;
    using namespace std;
    
    co int N=1e3+1,M=21;
    int n,m,a[N][M],c[M],ans;
    int head[N+M],edge[N*M*2],leng[N*M*2],next[N*M*2],tot;
    int d[N+M];
    
    il void add(int x,int y,int z){
    	edge[++tot]=y,leng[tot]=z,next[tot]=head[x],head[x]=tot;
    }
    bool bfs(){
    	memset(d,0,sizeof d),d[0]=1;
    	queue<int> q;q.push(0);
    	while(q.size()){
    		int x=q.front();q.pop();
    		for(int i=head[x];i;i=next[i]){
    			int y=edge[i],z=leng[i];
    			if(d[y]||!z) continue;
    			d[y]=d[x]+1,q.push(y);
    		}
    	}
    	return d[n+m+1]!=0;
    }
    int dinic(int x,int flow){
    	if(x==n+m+1) return flow;
    	int rest=flow;
    	for(int i=head[x];i&&rest;i=next[i]){
    		int y=edge[i],z=leng[i];
    		if(d[y]!=d[x]+1||!z) continue;
    		int delta=dinic(y,min(rest,z));
    		if(!delta) d[y]=0;
    		else{
    			leng[i]-=delta,leng[i^1]+=delta;
    			rest-=delta;
    		}
    	}
    	return flow-rest;
    }
    bool work(int l,int r){
    	tot=1;
    	for(int i=0;i<=n+m+1;++i) head[i]=0;
    	for(int i=1;i<=n;++i)for(int j=l;j<=r;++j)
    		add(i,n+a[i][j],1),add(n+a[i][j],i,0);
    	for(int i=1;i<=n;++i) add(0,i,1),add(i,0,0);
    	for(int i=1;i<=m;++i) add(n+i,n+m+1,c[i]),add(n+m+1,n+i,0);
    	int now=0,ans=0;
    	while(bfs())
    		while((now=dinic(0,0x7fffffff))) ans+=now;
    	return ans==n;
    }
    bool pd(int now){
    	for(int l=1;l+now-1<=m;++l)
    		if(work(l,l+now-1)) return 1;
    	return 0;
    }
    int main(){
    	read(n),read(m);
    	for(int i=1;i<=n;++i)for(int j=1;j<=m;++j) read(a[i][j]);
    	for(int i=1;i<=m;++i) read(c[i]);
    	int l=1,r=m;
    	while(l<r){
    		int mid=l+r>>1;
    		if(pd(mid)) r=mid;
    		else l=mid+1;
    	}
    	printf("%d
    ",l);
    	return 0;
    }
    
  • 相关阅读:
    Splay复习
    带权并查集复习-HDU3038
    罗素悖论-图灵停机问题
    数独解法c++实现
    状压DP NOI2001 炮兵阵地
    区间第k大数
    分块随笔T2
    分块感想
    webkit的高级属性
    设计模式
  • 原文地址:https://www.cnblogs.com/autoint/p/11050294.html
Copyright © 2011-2022 走看看