D - Maximizing Advertising

题目链接:https://cn.vjudge.net/contest/250168#problem/D

题目大意:给你一些点的坐标,这些点属于两个帮派,让你将这些点分进两个不能重叠的矩形中,问你最多两个矩形中不同帮派之和为多少?

具体思路:

将点分别按照x升序进行排序,按照y升序进行排序。横着扫描一遍,竖着扫描一遍,求出在扫描的过程中和的最大值。

AC代码;

#include<bits/stdc++.h>
using namespace std;
# define maxn 1000000+10
struct node
{
    int x,y;
    char cost;
} q[maxn];
bool cmp1(node t1,node t2)
{
    return t1.x<t2.x;
}
bool cmp2(node t1,node t2)
{
    return t1.y<t2.y;
}
int xb[maxn],xw[maxn],yb[maxn],yw[maxn];
void init()
{
    memset(xb,0,sizeof(xb));
    memset(xw,0,sizeof(xw));
    memset(yb,0,sizeof(yb));
    memset(yw,0,sizeof(yw));
}
int main()
{
    int n;
    scanf("%d",&n);
    getchar();
    for(int i=1; i<=n; i++)
    {
        scanf("%d %d %c",&q[i].x,&q[i].y,&q[i].cost);
    }
    sort(q+1,q+n+1,cmp1);
    for(int i=1; i<=n; i++)
    {
        if(q[i].cost=='b')
        {
            xb[i]++;
        }
        if(q[i].cost=='w')
        {
            xw[i]++;
        }
        xb[i]+=xb[i-1];
        xw[i]+=xw[i-1];//求x的前缀和
    }
    sort(q+1,q+n+1,cmp2);
    for(int i=1; i<=n; i++)
    {
        if(q[i].cost=='b')
        {
            yb[i]++;
        }
        if(q[i].cost=='w')
        {
            yw[i]++;
        }
        yb[i]+=yb[i-1];
        yw[i]+=yw[i-1];//求y的前缀和
    }
    int maxx=-1;
    for(int i=1; i<=n; i++)
    {
        int temp1=xb[i]+xw[n]-xw[i];
        int temp2=xw[i]+xb[n]-xb[i];
        maxx=max(maxx,max(temp1,temp2));
    }
    for(int i=1; i<=n; i++)
    {
        int temp1=yb[i]+yw[n]-yw[i];
        int temp2=yw[i]+yb[n]-yb[i];
        maxx=max(maxx,max(temp1,temp2));
    }
    printf("%d\n",maxx);
    return 0;
}

 

posted @ 2018-08-27 20:10  Let_Life_Stop  阅读(124)  评论(0编辑  收藏  举报