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UVALive 6656 Watching the Kangaroo --二分

题意:给你一些区间,再查询一些点,问这些点与所有区间形成的最小距离的最大值。最小距离定义为:如果点在区间内,那么最小距离为0,否则为min(pos-L[i],R[i]-pos)。

解法:当然要排个序,仔细想想会发现我们要找的区间的位置满足二分性质,即如果此时pos-L[mid] >= R[mid]-pos,那么我们要找的区间肯定是mid或大于mid,否则,我们要找的区间一定是mid即mid以下。二分找到即可。预处理时要把嵌套在别的区间里的区间忽略掉,因为外面那个区间一定比他更优。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define ll long long
using namespace std;
#define N 100007
#define M 22

struct node
{
    ll l,r;
}p[N],np[N];
ll L[N],R[N];
int tot;
ll pos;

int cmp(node ka,node kb)
{
    return ka.l < kb.l;
}

ll get(int mid)
{
    if(mid > tot || mid < 1)
        return 0;
    if(L[mid] > pos || R[mid] < pos)
        return 0;
    return min(pos-L[mid],R[mid]-pos);
}

int main()
{
    int t,cs = 1,n,m,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%lld%lld",&p[i].l,&p[i].r);
        sort(p+1,p+n+1,cmp);
        tot = 1;
        for(i=2;i<=n;i++)
        {
            if(p[i].r > p[tot].r)
            {
                tot++;
                p[tot] = p[i];
            }
        }
        for(i=1;i<=tot;i++)
        {
            L[i] = p[i].l;
            R[i] = p[i].r;
        }
        printf("Case %d:\n",cs++);
        while(m--)
        {
            int ans = 0;
            scanf("%lld",&pos);
            int low,high;
            low = 1,high = tot;
            while(low <= high)
            {
                int mid = (low+high)/2;
                if(R[mid]-pos <= pos-L[mid])
                    low = mid+1;
                else
                    high = mid-1;
            }
            printf("%lld\n",max(get(low),get(high)));
        }
    }
    return 0;
}
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posted @ 2014-08-20 10:17  whatbeg  阅读(258)  评论(0编辑  收藏  举报