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UVa 12715 Watching the Kangaroo(二分)

题意:n条线段(n <= 100000) (L<=R <= 1e9) ,m组询问(m <= 100000) 每次询问一个点的覆盖范围的最大值,一个点x对于一条包含其的线段,覆盖范围为min(x-L,R-x)

思路:考虑将线段一份为二,对于左边的那部分,以右端点排序,然后 二分找到右端点恰好满足的那个点为id,那么接下来要做的就是就是在[id,n]这个范围内找到L最小的那个点,可以通过求前缀最大来得到。那么左边最大距离为  seg[id].L-preLMax[id],右边最大的求法也是类似。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int maxn = 100000+10;
#define REP(_,a,b) for(int _ = (a); _ <= (b); _++)
struct seg{
    int L,R;
    seg(int L = 0,int R = 0):L(L),R(R){}
};
bool cmp1(seg a,seg b) {
    if(a.R != b.R)  return a.R < b.R;
    else return a.L < b.L;
}
bool cmp2(seg a,seg b) {
    if(a.L != b.L)  return a.L < b.L;
    else return a.R < b.R;
}
vector<seg>lft,rgt;

int preLMax[maxn],preRMax[maxn];
int n,m;

int main(){

    int ncase,T=1;
    cin >> ncase;
    while(ncase--) {
        lft.clear();
        rgt.clear();
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; i++) {
            int L,R;
            scanf("%d%d",&L,&R);
            int mid = (L+R)>>1;
            lft.push_back(seg(L,mid));
            rgt.push_back(seg(mid,R));
        }

        sort(lft.begin(),lft.end(),cmp1);
        preLMax[lft.size()] = 1e9;
        for(int i = lft.size()-1; i >= 0; i--) {
            preLMax[i] = min(preLMax[i+1],lft[i].L);
        }
        sort(rgt.begin(),rgt.end(),cmp2);
        preRMax[0] = rgt[0].R;
        for(int i = 1; i < rgt.size(); i++) {
            preRMax[i] = max(preRMax[i-1],rgt[i].R);
        }
        printf("Case %d:\n",T++);
        while(m--) {
            int x,ans=0;
            scanf("%d",&x);
            int L = 0,R = lft.size()-1;
            while(L <= R) {
                int mid = (L+R) >>1;
                if(lft[mid].R < x) {
                    L = mid+1;
                }else{
                    R = mid-1;
                }
            }
            ans = max(ans,x-preLMax[L]);
            L = 0,R = rgt.size()-1;
            while(L <= R) {
                int mid = (L+R) >>1;
                if(rgt[mid].L < x) {
                    L = mid+1;
                }else{
                    R = mid-1;
                }
            }
            ans = max(ans,preRMax[R]-x);
            ans = max(ans,0);
            printf("%d\n",ans);
        }
    }
    return 0;
}


UVa 12715 Watching the Kangaroo(二分)