您的位置:首页 > 理论基础 > 计算机网络

2017 ACM-ICPC 亚洲区(南宁赛区)网络赛 Overlapping Rectangles

2017-11-07 20:26 375 查看
题意:n个矩形,给出左下角的坐标和右上角的坐标,算出n个矩形所覆盖的面积

HDU 1542极其类似

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn=1005;
typedef long long ll;
struct LINE
{
ll  x, y_down, y_up;
int  flag;
bool operator<(const LINE &a)const
{
return  x<a.x;
}
}line[2*maxn];

struct TREE
{
ll  y_down, y_up;
ll  x;
int     cover; //用以表示加进线段树中的线段次数
bool    flag; //此标记用来表示是否有超元线段;为了处理方便加上去的
}tree[1000*maxn];

int     n;
ll  x1, y_1, x2, y2;
int     index=0;
ll  y[2*maxn];

void build(int i, int l, int r)
{
tree[i].x = -1; //-1表示该区间已经没有线段
tree[i].cover = 0; //表示该区间上有多少条线段;左边线段加进去则++,右边线段加进去则--
tree[i].y_down = y[l];
tree[i].y_up = y[r];
tree[i].flag = false;
if(l+1==r)
{
tree[i].flag = true; //flag==true表示达到了叶子节点
return;
}
int mid=(l+r)>>1;
build(2*i, l, mid);
build(2*i+1, mid, r);
}

ll insert(int i, ll x, ll l, ll r, int flag) //flag表示为左边还是右边
{
if (r<=tree[i].y_down || l>=tree[i].y_up)
return 0;
if (tree[i].flag)
{
if (tree[i].cover > 0) //递归到了叶子节点
{
ll temp_x = tree[i].x;
ll ans=(x-temp_x)*(tree[i].y_up - tree[i].y_down);
tree[i].x = x;   //定位上一次的x
tree[i].cover += flag;
return ans;
}
else
{
tree[i].cover += flag;
tree[i].x = x;
return 0;
}
}
ll ans1, ans2;
ans1 = insert(2*i, x, l, r, flag);
ans2 = insert(2*i+1, x, l, r, flag);
return ans1+ans2;
}

int main( )
{
// freopen("d:\\in.txt","r",stdin);
while (scanf("%d", &n)!=EOF&&n)
{
index = 1;
for (int i=1; i<=n; i++)
{
scanf("%lld%lld%lld%lld", &x1, &y_1, &x2, &y2);
y[index] = y_1;
line[index].x = x1;
line[index].y_down = y_1;
line[index].y_up = y2;
line[index].flag = 1; //1表示左边

index++;
y[index] = y2;
line[index].x = x2;
line[index].y_down = y_1;
line[index].y_up = y2;
line[index].flag = -1; //-1表示右边
index++;
}
sort(&y[1], &y[index]); //把所有的纵坐标按从小到大排序,把1写成了0,WA一次
sort(&line[1], &line[index]);
build(1, 1, index-1);
ll ans=0;
for (int i=1;i<index; i++)
{
ans += insert(1, line[i].x, line[i].y_down, line[i].y_up, line[i].flag);
}
printf("%lld\n",ans);
}
printf("*\n");
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  acm-icpc
相关文章推荐