HOJ1119/HDU1542 Atlantis HOJ1909/POJ1177 Picture

第一题：题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1542

题意：求多个长方形在平面上所覆盖的面积和。

扫描线。从下往上扫描，浮点数离散化处理。

思路参考：http://www.cnblogs.com/scau20110726/archive/2013/03/21/2972808.html

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

#define Maxn 250
#define lx (x<<1)
#define rx ((x<<1)|1)
#define MID ((l + r)>>1)

double X[Maxn];
double S[Maxn<<2];
int cnt[Maxn<<2];

struct Seg
{
    double l;
    double r;
    double h;
    int s;
    Seg(){}
    Seg(double _l,double _r,double _h,int _s)
    {
        l = _l;r = _r;h = _h;s = _s;
    }
    bool operator <(const Seg & a) const
    {
        return h<a.h;
    }
};
Seg seg[Maxn];

int binarySearch(int l,int r,double x)
{
    while(l<=r)
    {
        int mid = (l + r)>>1;
        if(X[mid] == x) return mid;
        if(X[mid]< x) l = mid+1;
        else r = mid-1;
    }
    return 0;
}
void pushUp(int l,int r,int x)
{
    if(cnt[x]) S[x] = X[r+1] - X[l];
    else if(l == r) S[x] = 0;
    else S[x] = S[lx] + S[rx];
}
void update(int L,int R,int d,int l,int r,int x)
{
    if(L<=l && r<=R)
    {
        cnt[x] += d;
        pushUp(l,r,x);
        return;
    }
    if(L<=MID) update(L,R,d,l,MID,lx);
    if(MID+1<=R) update(L,R,d,MID+1,r,rx);
    pushUp(l,r,x);
}
void init()
{
    memset(S,0,sizeof(S));
    memset(cnt,0,sizeof(cnt));
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    int cas = 0;
    int n;
    double a,b,c,d;
    int p = 0;
    while(scanf(" %d",&n)!=EOF && n!=0)
    {
        init();
        cas++;
        p = 0;
        for(int i=0;i<n;i++)
        {
            scanf(" %lf %lf %lf %lf",&a,&b,&c,&d);
            X[p] = a;
            seg[p++] = Seg(a,c,b,1);
            X[p] = c;
            seg[p++] = Seg(a,c,d,-1);
        }
        sort(X,X+p);
        sort(seg,seg+p);
        int k = 1;
        //去重
        for(int i=1;i<p;i++)
        {
            if(X[i]!=X[i-1]) X[k++] = X[i];
        }
        //从下往上扫描，忽略最上部边
        double ans = 0;
        for(int i=0;i<p-1;i++)
        {
            int l = binarySearch(0,k-1,seg[i].l);
            int r = binarySearch(0,k-1,seg[i].r) - 1;
            update(l,r,seg[i].s,0,k-1,1);
            ans += S[1] * (seg[i+1].h - seg[i].h);
        }
        printf("Test case #%d\nTotal explored area: %.2lf\n\n",cas,ans);
    }
    return 0;
}

第二题：题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1828

题意：求在平面内的长方形所形成的轮廓的周长。

思路和求面积相似。扫描线从下网上扫描，分为横边长度的记录len[]和竖边个数的记录numSeg[].分别将竖边长度和横边长度的改变绝对值都求出相加即可。

另外要考虑到竖边会重叠。使用lbd[]和rbd[]比对一下是否会重叠。

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

#define Maxn 15005
#define lx (x<<1)
#define rx ((x<<1)|1)
#define MID ((l + r)>>1)

int cnt[Maxn<<2];

//在横轴投影的有效长度
int len[Maxn<<2];
//在横轴投影的有效线段个数×2，即竖边的个数
int numSeg[Maxn<<2];

//区间左边界是否有竖边
bool lbd[Maxn<<2];
//区间右边界是否有竖边
bool rbd[Maxn<<2];

struct Seg
{
    int l,r,h,s;
    Seg(){}
    Seg(int _l,int _r,int _h,int _s)
    {
        l = _l;r = _r;h = _h;s = _s;
    }
    bool operator <(const Seg &a) const
    {
        return h<a.h || (h == a.h && s>a.s);
    }
};
Seg seg[Maxn];

void pushUp(int l,int r,int x)
{
    if(cnt[x])
    {
        len[x] = (r - l + 1);
        numSeg[x] = 2;
        lbd[x] = rbd[x] = 1;
    }
    else if(l == r)
    {
        len[x] = numSeg[x] = lbd[x] = rbd[x] = 0;
    }
    else
    {
        lbd[x] = lbd[lx];
        rbd[x] = rbd[rx];
        len[x] = len[lx] + len[rx];
        numSeg[x] = numSeg[lx] + numSeg[rx];
        if(lbd[rx] && rbd[lx]) numSeg[x] -= 2;
    }
}
void update(int L,int R,int d,int l,int r,int x)
{
    if(L<=l && r<=R)
    {
        cnt[x]+= d;
        pushUp(l,r,x);
        return;
    }
    if(L<=MID) update(L,R,d,l,MID,lx);
    if(MID+1<=R) update(L,R,d,MID+1,r,rx);
    pushUp(l,r,x);
}
void init()
{
    memset(cnt,0,sizeof(cnt));
    memset(numSeg,0,sizeof(numSeg));
    memset(len,0,sizeof(len));
    memset(lbd,0,sizeof(lbd));
    memset(rbd,0,sizeof(rbd));
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    int n;
    int a,b,c,d;
    int p,last = 0;
    while(scanf(" %d",&n)!=EOF)
    {
        init();
        p = 0;
        last = 0;
        int leftMin = 10005,rightMax = -10005;
        for(int i=0;i<n;i++)
        {
            scanf(" %d %d %d %d",&a,&b,&c,&d);
            if(a<leftMin) leftMin = a;
            if(c>rightMax) rightMax = c;
            seg[p++] = Seg(a,c,b,1);
            seg[p++] = Seg(a,c,d,-1);
        }
        sort(seg,seg+p);
        int ans = 0;
        for(int i=0;i<p;i++)
        {
            update(seg[i].l,seg[i].r-1,seg[i].s,leftMin,rightMax-1,1);
            //为什么要取绝对值，想一下
            //横边长度
            ans += abs(len[1] - last);
            last = len[1];
            //竖边长度
            ans += numSeg[1]*(seg[i+1].h-seg[i].h);
        }
        printf("%d\n",ans);
    }
    return 0;
}