題目鏈接：

D. Vika and Segments

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard 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.

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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.

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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).

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題意：

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給了這么多線段,問(wèn)它們一共包含了多少個(gè)點(diǎn);

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思路：

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把線段變成寬為1的矩形,然后用掃描線算法求面積;

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AC代碼:

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+4;
int n,x1,x2,y3,y2,rec[2*N],num;
struct no
{int l,r,h,flag;
};
no line[8*N];
struct nod
{int l,r,cover;ll sum;
};
nod tree[8*N];
int cmp(no x,no y)
{return x.h<y.h;
}
void build(int node,int L,int R)
{tree[node].l=L,tree[node].r=R;tree[node].cover=tree[node].sum=0;if(L>=R)return ;int mid=(L+R)>>1;build(2*node,L,mid);build(2*node+1,mid+1,R);
}
void Pushup(int node)
{if(tree[node].cover){tree[node].sum=rec[tree[node].r+1]-rec[tree[node].l];}else{if(tree[node].l==tree[node].r)tree[node].sum=0;else tree[node].sum=tree[2*node].sum+tree[2*node+1].sum;}
}
void update(int node,int L,int R,int x)
{if(L<=tree[node].l&&R>=tree[node].r){tree[node].cover+=x;Pushup(node);return ;}int mid=(tree[node].l+tree[node].r)>>1;if(L>mid) update(2*node+1,L,R,x);else if(R<=mid)update(2*node,L,R,x);else{update(2*node,L,mid,x);update(2*node+1,mid+1,R,x);}Pushup(node);
}
int bi(int x)
{int L=1,R=num-1,mid;while(L<=R){mid=(L+R)>>1;if(rec[mid]==x)return mid;else if(rec[mid]>x)R=mid-1;else L=mid+1;}return -1;
}
int main()
{scanf("%d",&n);int cnt=1;for(int i=1;i<=n;i++){scanf("%d%d%d%d",&x1,&y3,&x2,&y2);if(x1>x2)swap(x1,x2);if(y3>y2)swap(y3,y2);rec[cnt] = line[cnt].l = x1;line[cnt].r = x2+1;line[cnt].h = y3;line[cnt++].flag = 1;line[cnt].l = x1;rec[cnt] = line[cnt].r = x2+1;line[cnt].h = y2+1;line[cnt++].flag = -1;}sort(line+1,line+cnt,cmp);sort(rec+1,rec+cnt);num = 2;for(int i = 2;i < cnt;i++){if(rec[i]!=rec[i-1])rec[num++]=rec[i];}build(1,1,num-1);ll ans=0;for(int i = 1;i < cnt-1;i++){int fx = bi(line[i].l);int fy = bi(line[i].r)-1;if(fx <= fy){update(1,fx,fy,line[i].flag);}ans+=tree[1].sum*(ll)(line[i+1].h-line[i].h);}cout<<ans<<"\n";return 0;
}

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轉(zhuǎn)載于:https://www.cnblogs.com/zhangchengc919/p/5363831.html

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