ZOJ 1391 Horizontally Visible Segments

挺不错的一道线段树，薛神给的礼物，WA了整整一天才过，各种细节出错OTL……

1.纵坐标扩大两倍保存，不然样例都跑不过。即(0, 2)与(3, 4)之间可以有一条横线，但是如果坐标不扩大，这条横线是过不去的。

2.按x值从小到大排序，插入线段树，每次将Line[id]更新到线段树中之前，先查询编号为Line[id]向左可以看到哪几条线段（即可以看到哪几条编号比它小的线段），用vector保存，注意去重，不然最后结果会多。（vector忘了clear这里错了好几次。。。）

3.id[MAXN]用来标记该段最顶端的线段的编号，-1代表该段没被覆盖或者该段不是被同一条线段覆盖

4.四重循环暴力枚举，暂时也没想到别的好办法……或许是因为数据弱的缘故，交上去效率还可以，140ms AC。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;

const int MAXLEN = 8010 << 1;
const int MAXN = 8010;

struct MyLine
{
    int y1, y2;
    int x;
};

int N, bound;
vector<int> see[MAXN];
MyLine L[MAXN];

int  id[ MAXLEN << 2 ];

bool cmp( MyLine a, MyLine b )
{
    return a.x < b.x;
}

void PushUp( int rt )
{
    int lc = rt << 1;
    int rc = rt << 1 | 1;
    if ( id[lc] == id[rc] )
        id[rt] = id[lc];
    else
        id[rt] = -1;
    return;
}

void PushDown( int rt )
{
    int lc = rt << 1;
    int rc = rt << 1 | 1;
    if ( id[rt] != -1 )
        id[lc] = id[rc] = id[rt];

    return;
}

bool check( int c, int tar )   //去除重点
{
    int len = see[c].size();
    for ( int i = 0; i < len; ++i )
        if ( tar == see[c][i] ) return false;
    return true;
}

void Query( int L, int R, int c, int l, int r, int rt )
{
    if ( id[rt] != -1 )
    {
        if ( check( c, id[rt] ) ) see[c].push_back( id[rt] );
        return;
    }
    if ( l >= r ) return;
    PushDown(rt);
    int m = ( l + r ) >> 1;
    if ( L <= m ) Query( L, R, c, lson );
    if ( R > m ) Query( L, R, c, rson );

    PushUp(rt);
    return;
}

void Update( int L, int R, int c, int l, int r, int rt )
{
    if ( L <= l && r <= R )
    {
        id[rt] = c;
        return;
    }
    PushDown(rt);
    int m = ( l + r ) >> 1;
    if ( L <= m ) Update( L, R, c, lson );
    if ( R > m ) Update( L, R, c, rson );
    PushUp( rt );
    return;
}

int main()
{
    int T;
    // freopen( "in.txt", "r", stdin );
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d", &N );
        bound = 0;

        memset( id, -1, sizeof(id) );
        for ( int i = 0; i < N; ++i )
        {
            scanf("%d%d%d", &L[i].y1, &L[i].y2, &L[i].x );
            L[i].y1 *= 2;
            L[i].y2 *= 2;
            see[i].clear();  // 勿忘清空
            bound = max( bound, L[i].y2 );
        }

        sort( L, L + N, cmp );

        for ( int i = 0; i < N; ++i )
        {
            Query( L[i].y1, L[i].y2, i, 0, bound, 1 );
            Update( L[i].y1, L[i].y2, i, 0, bound, 1 );
        }

        int ans = 0;
        for ( int i = 0; i < N; ++i )
        {
            int sz = see[i].size();
            for ( int j = 0; j < sz; ++j )
            {
                int i2 = see[i][j];    //i can see i2
                int sz2 = see[i2].size();
                for ( int y = 0; y < sz2; ++y )  //i2 can see see[i2][y]
                    for ( int x = 0; x < sz; ++x )
                    {
                        if ( x != j && see[i][x] == see[i2][y] )
                            ++ans;
                    }
            }
        }

        printf( "%d\n", ans );
    }
    return 0;
}