POJ1436 线段树 区间更新

题中定义了什么叫做可见线段。所谓可见线段就是在给定的垂直与x轴的直线中，能够在两个直线之间连接一条平行线，并且这条平行线不与任何其他的直线相交。

WA一次，就是因为没有考虑到在建立点树的过程中会出现将 [1, 2] 和 [3, 4] 视为一条连接的直线，所以最后给坐标乘以一个2.

代码如下：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#define MAXN 8000
using namespace std;

struct Seq
{
    int y1, y2, x;
    bool operator < (Seq t) const
    {
        return x < t.x;
    }
}e[MAXN+5];

struct 
{
    int l, r, who;
}seg[MAXN*8];

map<int,int>m[MAXN+5];

void build(int f, int l, int r)
{
    int mid = l+r >> 1;
    seg[f].l = l, seg[f].r = r;
    seg[f].who = 0;
    if (r > l)
    {
        build(f<<1, l, mid);
        build(f<<1|1, mid+1, r);
    }
}

void up(int f)
{
    if (seg[f<<1].who == seg[f<<1|1].who)
        seg[f].who = seg[f<<1].who;
    else
        seg[f].who = -1;
}

void modify(int f, int l, int r, int who, map<int, int>*m)
{
    int mid = seg[f].l+seg[f].r >> 1;
    if (seg[f].l == l && seg[f].r == r && (seg[f].who != -1 || r==l))
    { 
        if (seg[f].who != -1 && seg[f].who != 0) 
            m[who][seg[f].who] = 1;
        seg[f].who = who;
            
    }
    else if (seg[f].r > seg[f].l)
    {
        if (seg[f].who == 0)
        { 
            seg[f<<1].who = seg[f<<1|1].who = seg[f].who;
            seg[f].who = -1;
        }
        else if (seg[f].who != -1)
        {
            m[who][seg[f].who] = 1;
            seg[f<<1].who = seg[f<<1|1].who = seg[f].who;
            seg[f].who = -1;
        }
        if (r <= mid)
            modify(f<<1, l, r, who, m);
        else if (l > mid)
            modify(f<<1|1, l, r, who, m);
        else
        {
            modify(f<<1, l, mid, who, m);
            modify(f<<1|1, mid+1, r, who, m);
        }
        up(f);
    }
}

int main()
{
    map<int,int>::iterator it1, it2;
    int T, M, y1, y2, x, ans;
    scanf("%d", &T);
    build(1, 0, 2*MAXN);
    while (T--)
    {
        ans = 0; 
        seg[1].who = 0;
        scanf("%d", &M);
        for (int i = 1; i <= M; ++i)
        {
            scanf("%d %d %d", &e[i].y1, &e[i].y2, &e[i].x);
            e[i].y1 <<= 1, e[i].y2 <<= 1;
            m[i].clear();
        }
        sort(e+1, e+1+M);
        for (int i = 1; i <= M; ++i)
        {
        //    printf("%d %d %d\n", e[i].y1, e[i].y2, e[i].x);
            modify(1, e[i].y1, e[i].y2, i, m);
        /*    for (it = m[i].begin(); it != m[i].end(); ++it)
                printf(".. %d ", it->first); */
        }
        for(int i = 1; i <= M; ++i) // 扫描整个表
        {
            for(it1 = m[i].begin(); it1 != m[i].end(); ++it1) // 遍历第i条线的可见线段  
            { 
                int k = it1->first;  // 记录与其中的第k条线段可见
                for(it2 = m[i].begin(); it2 != m[i].end(); ++it2)  // 如果第k条线段与第i条线段有共同的可见线段的话 ans++
                {
                    if (m[k].count(it2->first))
                        ans++;
                }
            }
        }
        printf("%d\n",ans);        
    }
    return 0;
}