【20.51%】【codeforces 610D】Vika and Segments

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black horizontal and vertical segments parallel to the coordinate axes. All segments have width equal to 1 square, that means every segment occupy some set of neighbouring squares situated in one row or one column.

Your task is to calculate the number of painted cells. If a cell was painted more than once, it should be calculated exactly once.

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of segments drawn by Vika.

Each of the next n lines contains four integers x1, y1, x2 and y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — the coordinates of the endpoints of the segments drawn by Vika. It is guaranteed that all the segments are parallel to coordinate axes. Segments may touch, overlap and even completely coincide.

Output
Print the number of cells painted by Vika. If a cell was painted more than once, it should be calculated exactly once in the answer.

Examples
input
3
0 1 2 1
1 4 1 2
0 3 2 3
output
8
input
4
-2 -1 2 -1
2 1 -2 1
-1 -2 -1 2
1 2 1 -2
output
16
Note
In the first sample Vika will paint squares (0, 1), (1, 1), (2, 1), (1, 2), (1, 3), (1, 4), (0, 3) and (2, 3).

【题目链接】:http://codeforces.com/contest/610/problem/D

【题解】

给你n条横线和纵线；
让你求这些线覆盖的点的面积(线上的一个点覆盖的面积为1);
最后面积不能重复;
做法:
只要把右上角的横纵坐标都加1;
就转化为扫描线求并矩形的面积问题了;
要把横坐标离散化下；
具体扫描线求并矩形面积请看这篇文章
http://blog.csdn.net/harlow_cheng/article/details/53027415

【完整代码】↓↓↓

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long

using namespace std;

const int MAXN = 1e5+100;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
struct abc
{
    LL l,r,h;
    int k;
};

int n,num = 0,cnt[MAXN<<3];
LL sum[MAXN<<3];
LL a1,b1,a2,b2;
abc bian[MAXN*2];
vector <LL> a;

void read2(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void read1(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

bool cmp(abc a,abc b)
{
    return a.h<b.h;
}

void push_up(int rt,int l,int r)
{
    if (cnt[rt])
        sum[rt]=a[r+1]-a[l];
    else
        if (l==r)
            sum[rt] = 0;
        else
            sum[rt] = sum[rt<<1]+sum[rt<<1|1];
}

void up_data(int L,int R,int c,int l,int r,int rt)
{
    if (L<=l && r<=R)
    {
        cnt[rt]+=c;
        push_up(rt,l,r);
        return;
    }
    int m = (l+r)>>1;
    if (L <= m)
        up_data(L,R,c,lson);
    if (m < R)
        up_data(L,R,c,rson);
    push_up(rt,l,r);
}

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    read1(n);
    for (int i = 1;i <= n;i++)
    {
        read2(a1);read2(b1);read2(a2);read2(b2);
        if (a1>a2)
            swap(a1,a2);
        if (b1>b2)
            swap(b1,b2);
        a2++;b2++;
        a.push_back(a1);a.push_back(a2);
        bian[++num].l = a1,bian[num].r = a2,bian[num].h = b1,bian[num].k = 1;
        bian[++num].l = a1,bian[num].r = a2,bian[num].h = b2,bian[num].k = -1;
    }
    sort(a.begin(),a.end());
    a.erase(unique(a.begin(),a.end()),a.end());
    sort(bian+1,bian+1+num,cmp);
    LL ans = 0;
    for (int i = 1;i <= num-1;i++)
    {
        int l = lower_bound(a.begin(),a.end(),bian[i].l)-a.begin();
        int r = lower_bound(a.begin(),a.end(),bian[i].r)-a.begin()-1;
        up_data(l,r,bian[i].k,0,a.size()-1,1);
        ans += sum[1]*(bian[i+1].h-bian[i].h);
    }
    cout << ans<<endl;
    return 0;
}