UVA221

这道题是紫书上的一道例题，在看刘汝佳老师的代码时对几个循环有些似懂非懂，后来从网上找到了一位大神的代码，讲解很仔细，想了一下就明白了，附上大神讲解的链接，希望有所帮助。点击打开链接

#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
const int maxn = 105;
int n;
double x[maxn * 2];
struct building
{
    int id;
    double x,y,w,d,h;
    bool operator < (const building& rhs)
    const
    {
        return x<rhs.x || (x==rhs.x&&y<rhs.y);
    }
} b[maxn];
bool cover(int i,double mx)
{
    return b[i].x<=mx&&b[i].x+b[i].w>=mx;
}
bool visable(int i,double mx)
{
    if(!cover(i,mx))
        return false;
    for(int k=0; k<n; k++)
        if(cover(k,mx)&&b[k].y<b[i].y&&b[k].h>=b[i].h)
            return false;
    return true;
}
int main()
{
    int kase = 0;
    while(cin>>n&&n)
    {
        for(int i=0; i<n; i++)
        {
            cin>>b[i].x>>b[i].y>>b[i].w>>b[i].d>>b[i].h;
            x[i*2] = b[i].x;
            x[i*2+1] = b[i].x + b[i].w;
            b[i].id = i + 1;
        }
        sort(b,b+n);
        sort(x,x+n*2);
        int m = unique(x,x+n*2) - x;
        if(kase++)
            cout<<endl;
        printf("For map #%d, the visible buildings are numbered as follows:\n%d",kase,b[0].id);
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<m-1; j++)
            {
                if(visable(i,(x[j]+x[j+1])/2))
                {
                    printf(" %d",b[i].id);
                    break;
                }
            }
        }
        cout<<endl;
    }
    return 0;
}