HDU 4302 Holedox Eating (二分+树状数组维护)
题意:一个小孩吃蛋糕,他的起始点在0位置,现在有m个操作,0 x 代表在x位置出现一个蛋糕; 1 代表小孩要去离他最近的点吃蛋糕,如果与他距离最小的点有两个,则去与他上次走的方向相同的点; 如果没有蛋糕,他就不动。 询问m次操作过后,小孩走了多少距离。
每一次,用二分判断,他右边离他最近的蛋糕点,和左边离他最近的蛋糕点。如何判断呢,只需用树状数组维护蛋糕总数,判断小孩当前位置pos 与 mid之间的蛋糕数是否大于等1,是的话可以尝试逼近,否则就远离。
需要注意线段是1 --- L ,而小孩起始点在0,所以有L + 1 个点。
#include <iostream>#include <algorithm>#include <cmath>#include<functional>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 100050#define INF 0x7FFFFFFFusing namespace std;int n,m;int c[MAX];int lowbit(int x) { return x & (-x);}void update(int x,int va) { while(x < n + 2) { c[x] += va; x += lowbit(x); }}int query(int x) { int sum = 0; while(x > 0) { sum += c[x]; x -= lowbit(x); } return sum;}int judge(int p,int rp,int lp,int flag) { if(rp == -1 && lp == -1) return -1; if(rp == -1) return 0; if(lp == -1) return 1; if(rp - p > p - lp) return 0; if(rp - p < p - lp) return 1; if(flag == 0) return 0; return 1;}int find(int pos,int l,int r,int kind) { int mid , des; if(kind == 1) des = INF; else des = -1; while(l <= r) { mid = (l + r) >> 1; if(kind == 1) { if(query(mid) - query(pos - 1) >= 1) { des = min(des,mid); r = mid - 1; } else l = mid + 1; } else { if(query(pos) - query(mid - 1) >= 1) { des = max(des,mid); l = mid + 1; } else r = mid - 1; }// cout << l << ' ' << mid << ' ' << r << endl; } if(des == INF) return -1; return des;}int main() { int T; cin >> T; int ca = 1; while(T--) { scanf("%d%d",&n,&m); memset(c,0,sizeof(c)); int a,b; int pos = 1, flag = 1; //flag 1:right 0: left __int64 dis = 0; for(int i=0; i<m; i++) { scanf("%d",&a); if(a == 1) { int rpos = find(pos,pos,n+1,1); int lpos = find(pos,1,pos,0); int dir = judge(pos,rpos,lpos,flag); if(dir == -1) continue; if(dir == 1) { dis += rpos - pos; pos = rpos; flag = 1; } else { dis += pos - lpos; pos = lpos; flag = 0; } update(pos,-1); } else { scanf("%d",&b); update(b+1,1); } } printf("Case %d: %d\n",ca ++,dis); } return 0;}