HDU 4614 Vases and Flowers （线段树[区间赋值+区间求和] + 二分）

题意：

给你n 个花瓶，每个花瓶只能放一朵花。两种操作

操作1：

给你f 朵花， 从a位置开始放，如果空花瓶不够了，或者花放完了， 就不放了。 输出 放的第一朵的花的位置 和最后一朵花的位置。

操作2：

给定区间[a,b] 将这个区间内的花瓶清空， 输出扔花的数量。

思路：

操作2很好处理，  先查询这个区间花的数量（有花可以设置为1，没花为0）， 在区间修改。 即为 区间赋值和区间求和的线段树。

操作1：

根据求区间和  二分第一个放花的位置pos1，  在根据pos1，二分最后一个放花的位置 pos2.

pos2 二分起来或许比较麻烦。

只要保证每次二分只查一次就好了， （查多了 可能会超时）

#include 
#include 
#include 
using namespace std;
int T;

const int maxn = 50000 + 10;
int setv[maxn<<2];
int sum[maxn<<2];

void pushdown(int o,int l,int r){

    if (setv[o] != -1){
        int m = l + r >> 1;
        int lson = o<<1;
        int rson = o<<1|1;
        setv[lson] = setv[rson] = setv[o];
        sum[lson] = setv[o] * (m-l+1);
        sum[rson] = setv[o] * (r-m);
        setv[o] = -1;
    }
}

void pushup(int o){
    sum[o] = sum[o<<1] + sum[o<<1|1];
}

int query(int L,int R,int l,int r,int o){
    if (L <= l && r <= R){
        return sum[o];
    }
    int m = l + r >> 1;
    pushdown(o,l,r);
    int ans = 0;
    if (m >= L){
        ans += query(L,R,l,m,o<<1);
    }
    if (m < R){
        ans += query(L,R,m+1,r,o<<1|1);
    }

    pushup(o);
    return ans;
}
void update(int L,int R,int c,int l,int r,int o){
    if (L <= l && r <= R){
        setv[o] = c;
        sum[o] = c * (r-l+1);
        return;
    }
    pushdown(o,l,r);
    int m = l + r >> 1;
    if (m >= L ){
        update(L,R, c,l,m,o<<1);
    }
    if (m < R){
        update(L,R,c,m+1,r,o<<1|1);
    }
    pushup(o);

}
int n,q;
void solve(int a,int f){
    int l = a,r = n-1;
    if (query(a,n,0,n-1,1) == n-a) {
        puts("Can not put any one.");
        return;
    }
    int pos1 = 1;
    while(l <= r){
        int m = l + r >> 1;
        int t = query(a,m,0,n-1,1);
        if (t == m-a+1) l = m + 1;
        else {
            r = m - 1;
        }
    }
    pos1 = l;
    l = pos1;
    r = n-1;
    int pos2 = 1;
    int t = query(pos1, n-1, 0,n-1,1);
    if (n-pos1 - t < f) f = n-pos1-t;
//    printf("f = %d\n", query(2,4,0,n-1,1));
    while(l <= r) {
        int m = l + r >> 1;
        int t = query(pos1, m, 0, n-1 , 1);
        if (m - pos1 + 1 - t > f) {
            r = m - 1;
//            pos2 = m;
        }
        else if (m - pos1 + 1 - t == f){
            r = m - 1;
            pos2 = m;
        }
        else {
            l = m + 1;
            pos2 = m;
        }
    }
    printf("%d %d\n", pos1,l);
    update(pos1,l,1,0,n-1,1);
}

int main(){
    scanf("%d",&T);
    while(T--){
        memset(setv,-1,sizeof setv);
        memset(sum,0,sizeof sum);

        scanf("%d %d", &n, &q);
        while(q--){
            int op;
            scanf("%d",&op);
            if (op == 1){
                int a,f;
                scanf("%d %d",&a, &f);
                solve(a,f);

            }
            else {
                int a,b;
                scanf("%d %d",&a, &b);
                printf("%d\n", query(a,b,0,n-1,1));
                update(a,b,0,0,n-1,1);
            }
        }
        puts("");
    }



    return 0;
}

Vases and Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3447    Accepted Submission(s): 1387

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

2 10 5 1 3 5 2 4 5 1 1 8 2 3 6 1 8 8 10 6 1 2 5 2 3 4 1 0 8 2 2 5 1 4 4 1 2 3

 

Sample Output

[pre]3 7 2 1 9 4 Can not put any one. 2 6 2 0 9 4 4 5 2 3 [/pre]

 

Author

SYSU

 

Source

2013 Multi-University Training Contest 2