hnu12884 Area Coverage 矩形分割 or 线段树

题意：给你n个二维平面上的矩形，可以两两覆盖，问最后覆盖的总面积为多少

解题思路：1）矩形状分割，可以知道，每多出一个矩形就和前面所有产生的矩形判断，看是有相交，如果有的话，就对前面的矩形进行分割，最多可以分割成8块，因为这个算法是n×n的算法时间复杂度，所以我们还需要在枚举的时候加速，采用引导值（下一个不为0的矩阵的值），最后6700ms过了

解题代码：

// File Name: rect1.c
// Author: darkdream
// Created Time: 2014年02月20日 星期四 20时32分02秒
/*
ID: dream.y1
PROG: rect1
LANG: C++
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#define LL long long 
struct node{
    int lx,ly,rx,ry;
}a[30000];
int A, B,k;
int isin(struct node temp,int  x,int y)
{
    if( x >= temp.lx && x <= temp.rx && y >= temp.ly && y <= temp.ry)
        return 1; 
    return 0 ; 
}
int t = 0;
LL isaren(int i)
{
    if(a[i].lx != -1)
       return 1ll*(a[i].rx - a[i].lx +1) * (a[i].ry - a[i].ly +1) ;
    else return 0 ; 
}
void fun0(struct node p ,struct node q)
{
    return ;
}
void fun1(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = q.lx ; 
    a[t].ly = p.ry +1; 
    a[t].rx = p.lx -1;
    a[t].ry = q.ry;
    if(isaren(t) <= 0)
        t--;
}
void fun2(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = p.lx ; 
    a[t].ly = p.ry +1; 
    a[t].rx = p.rx;
    a[t].ry = q.ry;
    if(isaren(t) <= 0)
        t--;

}
void fun3(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = p.rx +1 ; 
    a[t].ly = p.ry +1 ; 
    a[t].rx = q.rx;
    a[t].ry = q.ry;
    if(isaren(t) <= 0)
        t--;

}
void fun4(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = q.lx ; 
    a[t].ly = p.ly; 
    a[t].rx = p.lx-1;
    a[t].ry = p.ry;
    if(isaren(t) <= 0)
        t--;
}
void fun5(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = p.rx +1 ; 
    a[t].ly = p.ly; 
    a[t].rx = q.rx;
    a[t].ry = p.ry;
    if(isaren(t) <= 0)
        t--;

}
void fun6(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = q.lx ; 
    a[t].ly = q.ly; 
    a[t].rx = p.lx - 1;
    a[t].ry = p.ly - 1;
    if(isaren(t) <= 0)
        t--;
}
void fun7(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = p.lx ; 
    a[t].ly = q.ly; 
    a[t].rx = p.rx;
    a[t].ry = p.ly - 1;
    if(isaren(t) <= 0)
        t--;

}
void fun8(struct node p,struct node q)
{
    t ++ ; 
    a[t].lx = p.rx + 1 ; 
    a[t].ly = q.ly; 
    a[t].rx = q.rx;
    a[t].ry = p.ly -1;
    if(isaren(t) <= 0)
        t--;

}
LL ans = 0;
void figu()
{
    for(int  i = 1;i <= t;i ++)
    {
        if(isaren(i) >= 1)
        {
        //    printf("%d %d %d %d
",a[i].lx,a[i].ly,a[i].rx,a[i].ry);
            ans+= isaren(i);
        }
    }
    printf("%I64d
",ans);
}
int lastjudge(node tn1,node tn2)
{
    if(tn1.lx >= tn2.lx && tn1.lx <= tn2.rx && tn1.rx >= tn2.lx && tn1.rx <= tn2.rx && tn1.ly < tn2.ly && tn1.ry > tn2.ry)
        return 1; 
    if(tn1.ly >= tn2.ly && tn1.ly <= tn2.ry && tn1.ry >= tn2.ly && tn1.ry <= tn2.ry && tn1.lx < tn2.lx && tn1.rx > tn2.rx)
        return 1;
    return 0 ; 
}
int next[30004];
int main(){
    int ca;
    scanf("%d",&ca);
    while(ca--)
    {
        ans = 0 ; 
        t = 0 ; 
        scanf("%d",&k);
        for(int i = 1;i <= 30000;i ++)
            next[i] = i+1;
        for(int i = 1 ;i <= k;i ++)
        {
            t ++ ; 
            scanf("%d %d %d %d",&a[t].lx,&a[t].ly,&a[t].rx,&a[t].ry);
            a[t].rx -- ;
            a[t].ry --;
            int limit = t ;
            int tnext  = 1;  
            for(int j = 1;j < limit ;)
            {
                //printf("%d
",j);
                //printf("%d
",j);
                if(j == 0 )
                    break;
                struct node tn1,tn2;
                tn1 = a[limit];
                tn2 = a[j];
                if(isin(tn2,tn1.lx,tn1.ly) || isin(tn2,tn1.rx,tn1.ry) || isin(tn2,tn1.lx,tn1.ry) || isin(tn2,tn1.rx,tn1.ly)||isin(tn1,tn2.lx,tn2.ly) || isin(tn1,tn2.rx,tn2.ry) || isin(tn1,tn2.lx,tn2.ry) || isin(tn1,tn2.rx,tn2.ly)||lastjudge(tn1,tn2))
                {
                    a[j].lx = a[j].ly = a[j].rx = a[j].ry = -1; 
                    int hs[10];
                    memset(hs,0,sizeof(hs));
                    if(tn1.lx >= tn2.lx  && tn1.lx <= tn2.rx)
                    {
                        hs[6] = 1; 
                        hs[7] = 1; 
                        hs[8] = 1; 
                    }else tn1.lx = tn2.lx;

                    if(tn1.rx >= tn2.lx  && tn1.rx <= tn2.rx)
                    {
                        hs[1] = 1; 
                        hs[2] = 1; 
                        hs[3] = 1; 
                    }else tn1.rx = tn2.rx;

                    if(tn1.ly >= tn2.ly  && tn1.ly <= tn2.ry)
                    {
                        hs[1] = 1; 
                        hs[4] = 1; 
                        hs[6] = 1; 
                    }else tn1.ly = tn2.ly;

                    if(tn1.ry >= tn2.ly  && tn1.ry <= tn2.ry)
                    {
                        hs[3] = 1; 
                        hs[5] = 1; 
                        hs[8] = 1; 
                    }else tn1.ry = tn2.ry;
                    //     printf("(((((((%d %d %d %d %d
%d %d %d %d %d))))))
",tn1.lx,tn1.ly,tn1.rx,tn1.ry,tn1.c,tn2.lx,tn2.ly,tn2.rx,tn2.ry,tn2.c);
                    void (*p[])(node,node) = {fun0,fun1,fun2,fun3,fun4,fun5,fun6,fun7,fun8};
                    for(int s =1 ;s <= 8 ;s ++)
                    {
                        p[s](tn1,tn2);
                    }
                }
                if(isaren(j) > 0 && j != 1)
                {
                   next[tnext] = j ;
                   tnext = j;  
                }
                j = next[j];  
            }
        }
        figu();
    }
    return 0 ;
}

2)还可以用线段树来求解这题