zoj 1610 线段树

用线段树进行区间赋值，最后将每个小segment的颜色求出来，再扫一遍判断连续的段数即可。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 8001;
int color[N];
int ans[N];

struct Node 
{
    int l, r, c;
} node[N << 2];

void build( int i, int l, int r )
{
    node[i].l = l, node[i].r = r, node[i].c = -1;
    if ( l == r ) return ;
    int mid = ( l + r ) >> 1;
    build( i << 1, l, mid );
    build( i << 1 | 1, mid + 1, r );
}

void pushdown( int i )
{
    if ( node[i].c != -1 )
    {
        node[i << 1].c = node[i << 1 | 1].c = node[i].c;
        node[i].c = -1;
    }
}

void update( int i, int l, int r, int c )
{
    if ( node[i].l == l && node[i].r == r )
    {
        node[i].c = c;
        return ;
    }
    pushdown(i);
    int mid = ( node[i].l + node[i].r ) >> 1;
    if ( r <= mid )
    {
        update( i << 1, l, r, c );
    }    
    else if ( l > mid )
    {
        update( i << 1 | 1, l, r, c );
    }
    else
    {
        update( i << 1, l, mid, c );
        update( i << 1 | 1, mid + 1, r, c );
    }
}

void pushalldown( int i )
{
    if ( node[i].l == node[i].r )
    {
        if ( node[i].c != -1 )
        {
            color[node[i].l] = node[i].c;
        }
        return ;
    }
    pushdown(i);
    pushalldown( i << 1 );
    pushalldown( i << 1 | 1 );
}

int main ()
{
    int n;
    while ( scanf("%d", &n) != EOF )
    {
        build( 1, 1, 8000 );
        for ( int i = 1; i <= n; i++ )
        {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            update( 1, x + 1, y, z );
        }
        memset( color, -1, sizeof(color) );
        pushalldown(1);
        memset( ans, 0, sizeof(ans) );
        for ( int i = 1; i <= 8000; i++ )
        {
            if ( color[i] == -1 ) continue;
            ans[color[i]]++;
            while ( i + 1 <= 8000 && color[i] == color[i + 1] )
            {
                i++;
            }
        }
        for ( int i = 0; i <= 8000; i++ )
        {
            if ( ans[i] )
            {
                printf("%d %d
", i, ans[i]);
            }
        }
        putchar('
');
    }
    return 0;
}