HDU5699 2016"百度之星" - 初赛（Astar Round2B）货物运输

来源：互联网发布：程序员投资编辑：程序博客网时间：2024/04/29 09:14

二分答案判定

四条不等式简单推一下吧

#include <iostream>#include <cstdio>#include <cstring>#define INF 2147483647#define N 1000050using namespace std;int l[N],r[N],n,m;bool check(int x){    int mi1 = INF , mi2 = INF , ma1 = -INF , ma2 = -INF;    for (int i=1;i<=m;i++) if (r[i] - l[i] > x)     {        mi1 = min(mi1,r[i] + l[i] + x);        ma1 = max(ma1,r[i] + l[i] - x);        mi2 = min(mi2,r[i] - l[i] + x);        ma2 = max(ma2,r[i] - l[i] - x);    }    if(ma1 <= mi1 && ma2 <= mi2)      {          if(ma1 == mi1 && ma2 == mi2)          {              if((ma1 + ma2) & 1)                  return false;              return true;          }          return true;      }      return false;  }int main(){    while (~scanf("%d%d",&n,&m))    {        memset(l,0,sizeof(l));        memset(r,0,sizeof(r));        for (int i=1;i<=m;i++) {            scanf("%d%d",&l[i],&r[i]);            if (l[i] > r[i]) swap(l[i],r[i]);        }        int l = 0 , r = n;        while (l < r)        {            int mid = (l + r) / 2;            if (check(mid) == true) r = mid; else l = mid + 1;        }        printf("%d\n",l);    }    return 0;}

0 0