hdu4614 Vases and Flowers(简单线段树 + 二分)

Vases and Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1336    Accepted Submission(s): 532

Problem Description

　　Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.

 

Input

　　The first line contains an integer T, indicating the number of test cases.
　　For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).

 

Output

　　For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output 'Can not put any one.'. For each operation of which K is 2, output the number of discarded flowers. 
　　Output one blank line after each test case.

 

Sample Input

210 51 3 52 4 51 1 82 3 61 8 810 61 2 52 3 41 0 82 2 51 4 41 2 3

 

Sample Output

[pre]3 721 94Can not put any one.2 620 944 52 3[/pre]

 

Source

2013 Multi-University Training Contest 2

题目大意：n个花瓶，m个操作，花瓶里面有的有花，有的是空的。1操作是从a开始往右放b朵花，花瓶有了的不放，跳过，直到a右边都放满了花，多余的扔了。输出本次放花的起始位置，如果一朵不能放，输出一句话。

2操作是清除区间[a,b]的花。并输出清除了多少花。

题目分析：简单线段树。只需要维护区间和即可。对于1操作求起始位置，二分位置即可，对于操作2，统计求和即可。

详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>using namespace std;const int N = 50005;int n,m;struct node{    int sum,lazy;}tree[N<<2];void build(int num,int s,int e){    tree[num].sum = 0;    tree[num].lazy = -1;    if(s == e)        return;    int mid = (s + e)>>1;    build(num<<1,s,mid);    build(num<<1|1,mid + 1,e);}void pushup(int num){    tree[num].sum = tree[num<<1].sum + tree[num<<1|1].sum;}void pushdown(int num,int s,int e){    tree[num<<1].lazy = tree[num<<1|1].lazy = tree[num].lazy;    if(tree[num].lazy)    {        int mid = (s + e)>>1;        tree[num<<1].sum = mid - s + 1;        tree[num<<1|1].sum = e - mid;    }    else        tree[num<<1].sum = tree[num<<1|1].sum = 0;    tree[num].lazy = -1;}void update(int num,int s,int e,int l,int r,int val){    if(s == l && r == e)    {        tree[num].lazy = val;        if(val)            tree[num].sum = e - s + 1;        else            tree[num].sum = 0;        return;    }    if(tree[num].lazy > -1)        pushdown(num,s,e);    int mid = (s + e)>>1;    if(r <= mid)        update(num<<1,s,mid,l,r,val);    else    {        if(l > mid)            update(num<<1|1,mid + 1,e,l,r,val);        else        {            update(num<<1,s,mid,l,mid,val);            update(num<<1|1,mid + 1,e,mid + 1,r,val);        }    }    pushup(num);}int query(int num,int s,int e,int l,int r){    if(s == l && e == r)        return tree[num].sum;    if(tree[num].lazy > -1)        pushdown(num,s,e);    int mid = (s + e)>>1;    if(r <= mid)        return query(num<<1,s,mid,l,r);    else    {        if(l > mid)            return query(num<<1|1,mid + 1,e,l,r);        else            return query(num<<1,s,mid,l,mid) + query(num<<1|1,mid + 1,e,mid + 1,r);    }}int bin(int s,int rank){    int l = s;    int r = n;    int ans = -1;    int mid;    while(l <= r)    {        int mid = (l + r)>>1;        int tmp = query(1,1,n,s,mid);        if(tmp + rank == mid - s + 1)        {            ans = mid;            r = mid - 1;        }        else        {            if(tmp + rank < mid - s + 1)                r = mid - 1;            else                l = mid + 1;        }    }    return ans;}int main(){    int i,t;    int op,a,b;    scanf("%d",&t);    while(t --)    {        scanf("%d%d",&n,&m);        build(1,1,n);        while(m --)        {            scanf("%d%d%d",&op,&a,&b);            if(op == 1)            {                int st,ed;                a ++;                st = bin(a,1);                if(st == -1)                    puts("Can not put any one.");                else                {                    int tmp = query(1,1,n,st,n);                    tmp = n - st + 1 - tmp;                    if(tmp <= b)                        b = tmp;                    ed = bin(a,b);                    printf("%d %d\n",st - 1,ed - 1);                    update(1,1,n,st,ed,1);                }            }            else            {                a ++;b ++;                printf("%d\n",query(1,1,n,a,b));                update(1,1,n,a,b,0);            }        }        puts("");    }    return 0;}//890MS1256K