poj 3468 A Simple Problem with Integers( 线段树求区间和 )

从这题似乎感悟到了点线段树对区间操作的lazy思想，即只需要找到合适区间对其更新并标记，子区间暂不更新。通过画图模拟更加直观。

 

#include <iostream>using namespace std;const int Max = 100000;// 对象struct TreeNode {int l, r;long long c; // 增量long long sum; // 区间和}tree[Max*4];//-----------------线段树操作-------------------//void Build( const int l, const int r, const int id ){tree[id].l = l;tree[id].r = r;tree[id].c = 0;tree[id].sum = 0;if( r - l == 1 ) return;int mid = ( l + r ) >> 1;Build( l, mid, id * 2 );Build( mid, r, id * 2 + 1 );}void Update( const int l, const int r, const int v, const int id ){if( tree[id].l == l && tree[id].r == r ) {tree[id].sum += ( r - l ) * v; // 区间每个元素加vtree[id].c += v; // lazy思想,记录该区间的增量,供求子区间和时使用return;}if( tree[id].c != 0 ) { tree[ id*2 ].sum += ( tree[ id*2 ].r - tree[ id*2 ].l ) * tree[id].c;tree[ id*2+1 ].sum += ( tree[ id*2+1 ].r - tree[ id*2+1 ].l ) * tree[id].c;tree[ id*2 ].c += tree[id].c;tree[ id*2+1 ].c += tree[id].c;tree[id].c = 0;}int mid = ( tree[id].l + tree[id].r ) >> 1;if( r <= mid ) Update( l, r, v, id * 2 );else if( l >= mid ) Update( l, r, v, id * 2 + 1 );else {Update( l, mid, v, id * 2 );Update( mid, r, v, id * 2 + 1 );}// 合并tree[id].sum = tree[ id*2 ].sum + tree[ id*2+1 ].sum;}long long Query( const int l, const int r, const int id ){if( tree[id].l == l && tree[id].r == r ) {return tree[id].sum;}if( tree[id].c != 0 ) {tree[ id*2 ].sum += ( tree[ id*2 ].r - tree[ id*2 ].l ) * tree[id].c;tree[ id*2+1 ].sum += ( tree[ id*2+1 ].r - tree[ id*2+1 ].l ) * tree[id].c;tree[ id*2 ].c += tree[id].c;tree[ id*2+1 ].c += tree[id].c;tree[id].c = 0;}int mid = ( tree[id].l + tree[id].r ) >> 1;if( r <= mid ) return Query( l, r, id * 2 );else if( l >= mid ) return Query( l, r, id * 2 + 1 );elsereturn Query( l, mid, id * 2 ) + Query( mid, r, id * 2 + 1 );}//-----------------------------------------------------//int main(){int N, M, a, b, v;while( scanf( "%d%d", & N, & M ) == 2 ){Build( 1, N + 1, 1 );for( int i = 1; i <= N; ++ i ) {scanf( "%d", & v );Update( i, i + 1, v, 1 );}char ch;while( M-- ) {getchar();scanf( "%c%d%d", & ch, & a, & b );if( ch == 'Q' ) {long long sum = Query( a, b + 1, 1 );printf( "%lld/n", sum );}else {scanf( "%d", & v );Update( a, b + 1, v, 1 );}}}system( "pause" );return 0;}