HDU 4614 Vases and Flowers

选段树区间更新

题目描述:插花，每个花瓶只允许插入一枝花。插花操作是从第A个花瓶开始插花，如果第A个花瓶空着，就插入一枝花，否则判断第Ａ＋１个花瓶，知道插完全部的Ｆ只花，或者是没有花瓶可插，把剩余的花弃掉，操作终止。输出插花的第一个花瓶和最后一个花瓶，如果一枝花都无法插入，那输出Can not put any one.。清理操作：清理区间［Ａ，Ｂ］内的话，输出清理的花的数目。

解题分析：线段区间内维护三个量，区间内插的花的数目，区间内第一个空花瓶，最后一个空花瓶。模拟即可。

代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 50000 + 10;struct Node{    int l,r;    int add,sum,first,last;}P[maxn << 2];int n,m;int tol;void build(int l,int r,int pos){    P[pos].l = P[pos].first = l;    P[pos].r = P[pos].last = r;    P[pos].add = -1;    P[pos].sum = 0;    if(l == r) return ;    int mid = (l + r) / 2;    build(l,mid,pos << 1);    build(mid + 1,r,pos << 1 | 1);}void pushup(int pos){    P[pos].sum = P[pos << 1].sum + P[pos << 1 | 1].sum;    if( P[pos].sum == (P[pos].r - P[pos].l + 1) )    {        P[pos].first = INF;        P[pos].last = -INF;    }    else    {        P[pos].first = min( P[pos << 1].first,P[pos << 1 | 1].first );        P[pos].last = max(P[pos << 1].last , P[pos << 1 | 1].last );    }}void pushdown(int pos){    if(P[pos].add != -1)    {        P[pos << 1].add = P[pos << 1 | 1].add = P[pos].add;        if(P[pos].add == 1)        {             P[pos << 1].sum = P[pos << 1].r - P[pos << 1].l + 1;             P[pos << 1].first = INF;             P[pos << 1].last = - INF;             P[pos << 1 | 1].sum = P[pos << 1 | 1].r - P[pos << 1 | 1].l + 1;             P[pos << 1 | 1].first = INF;             P[pos << 1 | 1].last = -INF;        }        else if(P[pos].add == 0)        {            P[pos << 1].sum = P[pos << 1 | 1].sum = 0;            P[pos << 1].first = P[pos << 1].l;            P[pos << 1].last = P[pos << 1].r;            P[pos << 1 | 1].first = P[pos << 1 | 1].l;            P[pos << 1 | 1].last = P[pos << 1 | 1].r;        }        P[pos].add = -1;    }}void update1(int L,int R,int pos,int &first,int &last){    if(tol <= 0) return;    if(P[pos].sum == (P[pos].r - P[pos].l + 1)) return;    int rest = (P[pos].r - P[pos].l + 1 ) - P[pos].sum;    if(L <= P[pos].l && P[pos].r <= R && rest <= tol)    {        first = min(first,P[pos].first);        last = max(last,P[pos].last);        tol -= rest;        P[pos].sum = P[pos].r - P[pos].l + 1;        P[pos].first = INF;        P[pos].last = -INF;        P[pos].add = 1;        return ;    }    pushdown(pos);    int mid = (P[pos].l + P[pos].r) / 2;    if(L <= mid && tol > 0) update1(L,R,pos << 1,first,last);    if(mid < R && tol > 0) update1(L,R,pos << 1 | 1,first,last);    pushup(pos);}int update2(int L,int R,int pos){    if(L <= P[pos].l && P[pos].r <= R)    {        int t = P[pos].sum;        P[pos].sum = 0;        P[pos].first = P[pos].l;        P[pos].last = P[pos].r;        P[pos].add = 0;        return t;    }    pushdown(pos);    int mid = ( P[pos]. l + P[pos].r ) / 2;    int ans = 0;    if(L <= mid) ans += update2(L,R,pos << 1);    if(mid < R) ans += update2(L,R,pos << 1 | 1);    pushup(pos);    return ans;}int main(){    int kase;    scanf("%d",&kase);    while(kase--)    {        scanf("%d %d",&n,&m);        build(1,n,1);        while(m--)        {            int k,x,y;            scanf("%d %d %d",&k,&x,&y);            if(k == 1)            {                int first = INF,last = -INF;                tol = y;                x++;                update1(x,n,1,first,last);                if(first == INF) printf("Can not put any one.\n");                else printf("%d %d\n",first - 1,last - 1);            }            else            {                x++;                y++;                cout << update2(x,y,1) << endl;            }        }        puts("");    }    return 0;}