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EOJ 2525 Light Switching

/*
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EOJ 2525 Light Switching
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----問(wèn)題描述：
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Farmer John tries to keep the cows sharp by letting them play with intellectual toys.
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One of the larger toys is the lights in the barn.
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Each of the N (2 <= N <= 100,000) cow stalls conveniently numbered 1..N has a colorful light above it.
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At the beginning of the evening, all the lights are off. The cows control the lights with a set of N pushbutton switches that toggle the lights; pushing switch i changes the state of light i from off to on or from on to off.
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The cows read and execute a list of M (1 <= M <= 100,000) operations expressed as one of two integers (0 <= operation <= 1).
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The first operation (denoted by a 0 command) includes two subsequent integers S_i and E_i (1 <= S_i <= E_i <= N) that indicate a starting switch and ending switch. They execute the operation by pushing each pushbutton from S_i through E_i inclusive exactly once.
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The second operation (denoted by a 1 command) asks the cows to count how many lights are on in the range given by two integers S_i and E_i (1 <= S_i <= E_i <= N) which specify the inclusive range in which the cows should count the number of lights that are on.
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Help FJ ensure the cows are getting the correct answer by processing the list and producing the proper counts.
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----輸入：
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* Line 1: Two space-separated integers: N and M
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* Lines 2..M+1: Each line represents an operation with three space-separated integers: operation, S_i, and E_i
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----輸出：
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* Lines 1..number of queries: For each output query, print the count as an integer by itself on a single line.
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----樣例輸入：
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4 5
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0 1 2
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0 2 4
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1 2 3
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0 2 4
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1 1 4
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INPUT DETAILS:
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Four lights; five commands. Here is the sequence that should be processed:
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=========Lights
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=========1 2 3 4
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Init:====O O O O , O = off * = on
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0 1 2 -->* * O O ,toggle lights 1 and 2
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0 2 4 -->* O * *
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1 2 3 -->1 ,count the number of lit lights in range 2..3
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0 2 4 -->* * O O ,toggle lights 2, 3, and 4
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1 1 4 -->2 ,count the number of lit lights in the range 1..4
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----樣例輸出：
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1
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2
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----分析：
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*/
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template<unsigned int N>
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class CProblem
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{
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public :
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void init( int b, int e ){
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init( 1, b, e );
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}
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int query( int b, int e ){
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begin = b;
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end = e;
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return query( 1 );
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}
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void modify( int b, int e ){
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begin = b;
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end = e;
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modify( 1 );
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}
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private :
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void init( int node, int b, int e ){
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left [ node ] = b;
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right [ node ] = e;
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sum [ node ] = 0;
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modified[ node ] = false;
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if( b + 1 < e ){
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init( node + node, b, ( b + e ) / 2 );
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init( node + node + 1, ( b + e ) / 2, e );
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}
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}
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int query( int node ){
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if( ( end <= left[ node ] ) || ( right[ node ] <= begin ) ){
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return 0;
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}
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if( ( begin <= left[ node ] ) && ( right[ node ] <= end ) ){
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return sum[ node ];
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}
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int len = ( right[ node ] > end ? end : right[ node ] ) - ( left[ node ] < begin ? begin : left[ node ] );
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return ( modified[ node ] ) ? ( len - query( node + node ) - query( node + node + 1 ) ) : ( query( node + node ) + query( node + node + 1 ) );
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}
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void modify( int node ){
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if( ( end <= left[ node ] ) || ( right[ node ] <= begin ) ){
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return;
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}
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if( ( begin <= left[ node ] ) && ( right[ node ] <= end ) ){
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sum[ node ] = right[ node ] - left[ node ] - sum[ node ];
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modified[ node ] = ! modified[ node ];
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return;
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}
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modify( node + node );
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modify( node + node + 1 );
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sum[ node ] = ( modified[ node ] ) ? ( right[ node ] - left[ node ] - sum[ node + node ] - sum[ node + node + 1 ] ) : ( sum[ node + node ] + sum[ node + node + 1 ] );
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}
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int left[ N + N ], right[ N + N ], sum[ N + N ], begin, end;
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bool modified[ N + N ];
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};
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#include <iostream>
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#include <cstdio>
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using namespace std;
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CProblem<150003> prob;
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int main(){
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int n, m, cmd, a, b;
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scanf( "%d%d", &n, &m );
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prob.init( 1, n + 1 );
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while( m-- ){
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scanf( "%d%d%d", &cmd, &a, &b );
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if( cmd ){
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printf( "%d\n", prob.query( a, b + 1 ) );
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}
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else{
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prob.modify( a, b + 1 );
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}
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}
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return 0;
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}
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posted on 2012-04-22 22:46 coreBugZJ 閱讀(645) 評(píng)論(0) 編輯收藏引用所屬分類: ACM 、Algorithm 、DataStructure 、課內(nèi)作業(yè)

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EOJ 2525 Light Switching