[线段树] HDU 1698

来源：互联网发布：2017年最流行的网络语编辑：程序博客网时间：2024/05/20 12:49

一步一步理解线段树： http://www.cnblogs.com/TenosDoIt/p/3453089.html#f

还是这个地址，这次用到了优化所用的区间延迟更新，时间复杂度是O(lgn)

PS：领教到了不好好命名变量的坑爹之处。。。。。。。

#include <cstdio>#include <cstring>#include <iostream>#define lson root * 2#define rson root * 2 + 1using namespace std;const int X = 100002 << 2;int Tree[ X ];int flag[ X ]; //延迟标记void build ( int root, int ST, int END ) {        if ( ST == END ) {                Tree[ root ] = 1;                flag[ root ] = 0;        } else {                int mid = ( ST + END ) / 2;                build ( lson, ST, mid );                build ( rson, mid + 1, END );                Tree[ root ] = Tree[ lson ] + Tree[ rson ];        }}// 延迟更新 时间O(lgn)// 当前root的标记向下更新void pushDown ( int root, int st, int end ) {        if ( flag[ root ] ) {                flag[ lson ] = flag[ rson ] = flag[ root ]; //传递z（要修改成）的值给子节点                int mid = ( st + end ) / 2;                Tree[ lson ] =                    ( mid - st + 1 ) * flag[ root ]; //根代表的和 直接等于这段区间大小 * z                Tree[ rson ] = ( end - ( mid + 1 ) + 1 ) * flag[ root ];                flag[ root ] = 0;        }}// ST END --> required// st,end --> nowvoid update ( int root, int st, int end, int ST, int END, int val ) {        if ( st > END || end < ST )                return;        //找到要更新的整段区间        if ( st >= ST && end <= END ) {                flag[ root ] = val;                Tree[ root ] = val * ( end - st + 1 ); //更新这一段的根节点总和                return;        }        //建立flag，延迟更新子节点        pushDown ( root, st, end );        int mid = ( st + end ) / 2;        if ( mid >= ST )                update ( lson, st, mid, ST, END, val );        if ( mid < END )                update ( rson, mid + 1, end, ST, END, val );        Tree[ root ] = Tree[ lson ] + Tree[ rson ];}int main () {        int n;        scanf ( "%d", &n );        for ( int i = 1; i <= n; ++i ) {                memset ( flag, 0, sizeof ( flag ) );                int N, Q;                scanf ( "%d%d", &N, &Q );                build ( 1, 1, N );                while ( Q-- ) {                        int x, y, z;                        scanf ( "%d%d%d", &x, &y, &z );                        update ( 1, 1, N, x, y, z );                }                printf ( "Case %d: The total value of the hook is %d.\n", i, Tree[ 1 ] );        }        return 0;}

阅读全文

0 0