两种做法。
第一种:标记区间最大值和最小值,若区间最小值>=P,则本区间+2c,若区间最大值<P,则本区间+c。非常简单的区间更新。
最后发一点牢骚:最后query查一遍就行,我这个2B竟然写了个for循环每个点查了一遍orz……然后比赛的时候就一直TLE还查不出原因……感觉线段树对我就像个诅咒一样,每场必不出,不管是多么简单的线段树,都会错在千奇百怪的地方……说到底也不过是对线段树掌握的不扎实罢了,sigh……以后要多加练习!
#include <cstdio> #include <cstring> #include <cstdlib> #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define lc rt << 1 #define rc rt << 1 | 1 using namespace std; const int MAXN = 200100; int N, M, P; int sum[MAXN << 2]; int maxi[MAXN << 2]; int mini[MAXN << 2]; int lazy[MAXN << 2]; void build( int l, int r, int rt ) { sum[rt] = lazy[rt] = 0; maxi[rt] = 0; mini[rt] = 0; if ( l == r ) return; int m = ( l + r ) >> 1; build( lson ); build( rson ); return; } inline void PushDown( int rt, int m ) { if ( lazy[rt] ) { lazy[lc] += lazy[rt]; lazy[rc] += lazy[rt]; sum[lc] += lazy[rt]*(m - (m >> 1) ); sum[rc] += lazy[rt]*(m >> 1); maxi[lc] += lazy[rt], mini[lc] += lazy[rt]; maxi[rc] += lazy[rt], mini[rc] += lazy[rt]; lazy[rt] = 0; } return; } inline void PushUp( int rt ) { sum[rt] = sum[lc] + sum[rc]; maxi[rt] = maxi[lc] > maxi[rc] ? maxi[lc] : maxi[rc]; mini[rt] = mini[lc] < mini[rc] ? mini[lc] : mini[rc]; return; } inline void update( int L, int R, int val, int l, int r, int rt ) { if ( L <= l && r <= R ) { if ( maxi[rt] < P ) { lazy[rt] += val; sum[rt] += val*(r - l + 1); maxi[rt] += val; mini[rt] += val; return; } else if ( mini[rt] >= P ) { lazy[rt] += 2*val; sum[rt] += 2*val*(r - l + 1); maxi[rt] += 2*val; mini[rt] += 2*val; return; } } if ( l == r ) return; PushDown( rt, r - l + 1 ); int m = ( l + r ) >> 1; if ( L <= m ) update( L, R, val, lson ); if ( R > m ) update( L, R, val, rson ); PushUp( rt ); return; } bool first; void query( int l, int r, int rt ) { if ( l == r ) { if ( first ) putchar(' '); first = true; printf( "%d", sum[rt] ); return; } PushDown( rt, r - l + 1 ); int m = ( l + r ) >> 1; query( lson ); query( rson ); return; } int main() { while ( scanf( "%d%d%d", &N, &M, &P ) == 3 ) { build( 1, N, 1 ); for ( int i = 0; i < M; ++i ) { int a, b, c; scanf( "%d%d%d", &a, &b, &c ); update( a, b, c, 1, N, 1 ); } first = false; query( 1, N, 1 ); puts(""); } return 0; }
第二种做法:线段树的特殊懒惰标记,方法跟 HDU 3954 一样。代码稍微改改就行。
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define lc rt << 1 #define rc rt << 1 | 1 using namespace std; const int MAXN = 200022; const int INF = 1 << 30; struct node { int exp, level; int min_dis; int flag; }; int N, M, P; int K; node Tr[ MAXN << 2 ]; int sum[20]; void build( int l, int r, int rt ) { Tr[rt].exp = Tr[rt].flag = 0; Tr[rt].level = 1; Tr[rt].min_dis = sum[1]; if ( l == r ) return ; int m = ( l + r ) >> 1; build( lson ); build( rson ); return; } void PushDown( int rt ) { if ( Tr[rt].flag ) { Tr[lc].exp += Tr[rt].flag * Tr[lc].level; Tr[lc].min_dis -= Tr[rt].flag; Tr[lc].flag += Tr[rt].flag; Tr[rc].exp += Tr[rt].flag * Tr[rc].level; Tr[rc].min_dis -= Tr[rt].flag; Tr[rc].flag += Tr[rt].flag; Tr[rt].flag = 0; } return; } void PushUp( int rt ) { Tr[rt].level = max( Tr[lc].level, Tr[rc].level ); Tr[rt].exp = max( Tr[lc].exp, Tr[rc].exp ); Tr[rt].min_dis = min( Tr[lc].min_dis, Tr[rc].min_dis ); return; } void Update( int L, int R, int v, int l, int r, int rt ) { if ( l == r ) { Tr[rt].exp += Tr[rt].level * v; while ( Tr[rt].exp >= sum[ Tr[rt].level ] ) ++Tr[rt].level; Tr[rt].min_dis = ( sum[ Tr[rt].level ] - Tr[rt].exp ) / Tr[rt].level; if( ( sum[ Tr[rt].level ] - Tr[rt].exp ) % Tr[rt].level ) ++Tr[rt].min_dis; return; } int m = ( l + r ) >> 1; if ( L == l && r == R ) { if ( v >= Tr[rt].min_dis ) { PushDown(rt); if ( R <= m ) Update( L, R, v, lson ); else if ( L > m ) Update( L, R, v, rson ); else { Update( L, m, v, lson ); Update( m + 1, R, v, rson ); } PushUp(rt); } else { Tr[rt].exp += Tr[rt].level * v; Tr[rt].min_dis -= v; Tr[rt].flag += v; } return; } PushDown(rt); if ( R <= m ) Update( L, R, v, lson ); else if ( L > m ) Update( L, R, v, rson ); else { Update( L, m, v, lson ); Update( m + 1, R, v, rson ); } PushUp(rt); return; } bool first; void Query( int l, int r, int rt ) { if ( l == r ) { if ( first ) putchar(' '); first = true; printf( "%d", Tr[rt].exp ); return; } PushDown(rt); int m = ( l + r ) >> 1; Query( lson ); Query( rson ); return; } int main() { K = 2; while ( scanf( "%d%d%d", &N, &M, &P ) == 3 ) { sum[1] = P; sum[2] = INF; build( 1, N, 1 ); while ( M-- ) { int a, b, c; scanf( "%d%d%d", &a, &b, &c ); Update( a, b, c, 1, N, 1 ); } first = false; Query( 1, N, 1 ); puts(""); } return 0; }