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

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;}