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  • POJ 1990 MooFest

    MooFest
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 5339   Accepted: 2300

    Description

    Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of course, mooing. When the cows all stand in line for a particular event, they moo so loudly that the roar is practically deafening. After participating in this event year after year, some of the cows have in fact lost a bit of their hearing. 

    Each cow i has an associated "hearing" threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they must speak at a volume level equal to the distance between them times max(v(i),v(j)). 

    Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume. 

    Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows. 

    Input

    * Line 1: A single integer, N 

    * Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location. 

    Output

    * Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows. 

    Sample Input

    4
    3 1
    2 5
    2 6
    4 3
    

    Sample Output

    57
    

    Source

    解题思路:首先按牛的声音大小进行排序,这样就保证了与当前牛交流的牛都是以当前牛的声音为标准的,然后考虑当前牛的左右两側的牛到当前牛的距离之和,令s1表示牛左側的牛的数目,s2表示牛的左側的牛到当前牛的距离之和,total表示当前的全部牛的总距离长度
    左側距离L=cow[i].x*s1-s2;右側距离R=(total-s2)-(i-s1)*cow[i].x;
    s1,s2用树状数组维护
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int cnt[20005],len[20005];
    struct COW
    {
    	int v;
    	int x;
    }cow[20005];
    int cmp(COW a,COW b)
    {
    	return a.v<b.v;
    }
    int lowbit(int x)
    {return x&(-x);}
    void add(int *s,int pos,int n,int val)
    {
    	int i;
    	for(i=pos;i<=n;i+=lowbit(i))
    		s[i]+=val;
    }
    int sum(int *s,int pos,int n)
    {
    	int i,sum=0;
    	for(i=pos;i>0;i-=lowbit(i))
    		sum+=s[i];
    	return sum;
    }
    int main()
    {
    	int i,n,m;
    	long long ans,total,s1,s2;
    	while(scanf("%d",&n)!=-1)
    	{
    		ans=total=m=0;
    		memset(cnt,0,sizeof(cnt));
    		memset(len,0,sizeof(len));
    		for(i=1;i<=n;i++)
    		{
    			scanf("%d%d",&cow[i].v,&cow[i].x);
    			if(m<cow[i].x)
    				m=cow[i].x;
    		}
    		sort(cow+1,cow+1+n,cmp);
    		for(i=1;i<=n;i++)
    		{
    			add(cnt,cow[i].x,m,1);
    			add(len,cow[i].x,m,cow[i].x);
    			total+=cow[i].x;
    			s1=sum(cnt,cow[i].x-1,m);
    			s2=sum(len,cow[i].x-1,m);
                            long long L=<strong style="font-family: 'Times New Roman', Times, serif; background-color: rgb(255, 255, 255);"></strong><pre name="code" class="cpp" style="display: inline !important;">s1*cow[i].x-s2;

    long long R=
    total-s2-(i-s1)*cow[i].x;

    ans+=cow[i].v*(L+R);}cout<<ans<<endl;}return 0;}
    
    

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  • 原文地址:https://www.cnblogs.com/zsychanpin/p/6882813.html
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