17 Cows POJ

Cows POJ - 2481

1 题意

  给出n个牛的取食范围 [Si,Ei],若Si<= Sj,Ej<=Ei,Ei-Si > Ej-Sj;则称i牛大于j牛,问对于每一个牛,有多少个牛比它大

2 分析

对于n个牛,先按照有断点进行排序,排序之后用树状数组插入和查询,只需要比较左端点,左端点比i牛小,说明比i牛大(相等的除外)

4 参考代码

const int LEN = 1e5+100;int tree[LEN];struct Cow{    int s,e;    int ID;};bool operator<(const Cow  &a,const Cow &b){    if(a.e!=b.e)        return a.e>b.e;    return a.s < b.s;}//int sum_of_len[LEN];Cow cow[LEN];int num[LEN];int maxn = 0;void Update(int x){    while(x<=maxn+1)    {        tree[x]++;        x += lowbit(x);    }}int Query(int x){    int sum = 0;    while(x>0)    {        sum += tree[x];        x -= lowbit(x);    }    return sum;}int main(){    //  freopen("D:\\in.txt","r",stdin);    int N;    while(cin>>N&&N)    {        maxn = 0;        memset(tree,0,sizeof(tree));        memset(num,0,sizeof(num));        for(int i = 1; i <= N; ++i)        {            scanf("%d %d",&cow[i].s,&cow[i].e);            cow[i].ID = i;            maxn = max(cow[i].s,maxn);        }        sort(cow+1,cow+N+1);        for(int i = 1; i <= N; ++i)        {            if (cow[i].e==cow[i-1].e&&cow[i].s==cow[i-1].s)                num[cow[i].ID] = num[cow[i-1].ID];            else            {                int t = 0;                t = Query(cow[i].s+1);                num[cow[i].ID] =t;            }            Update(cow[i].s+1);        }        for(int i = 1; i <= N; ++i)        {            if(i>1)                printf(" ");            printf("%d",num[i]);        }        printf("\n");    }    return 0;}