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线段树的区间合并,其中lsum代表区间左边最大连续长度,rsum代表区间最大连续长度,msum代表区间

最大连续长度。因为本题是查询连续区间的左端点,那么如果左儿子满足条件,就继续查询左儿子,左儿

子不满足条件,就判断左儿子的rsum + 右儿子的lsum是不是满足条件,最后才查询右儿子。

 

/*Accepted    2236K    594MS    C++    2574B    2012-08-21 12:15:23*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

const int MAXN = 50050;
int msum[MAXN << 2], lsum[MAXN << 2], rsum[MAXN << 2];
int col[MAXN << 2];

void PushDown(int l, int r, int rt)
{
    int mid = l + r >> 1;
    if(col[rt] != -1)
    {
        col[rt << 1] = col[rt << 1 | 1] = col[rt];
        msum[rt << 1] = lsum[rt << 1] = rsum[rt << 1] = col[rt] ? 0 : mid - l + 1;
        msum[rt << 1 | 1] = lsum[rt << 1 | 1] = rsum[rt << 1 | 1]
        = col[rt] ? 0 : r - mid;
        col[rt] = -1;
    }
}

void PushUp(int l, int r, int rt)
{
    int mid = l + r >> 1;
    lsum[rt] = lsum[rt << 1];
    rsum[rt] = rsum[rt << 1 | 1];
    if(lsum[rt] == mid - l + 1) lsum[rt] += lsum[rt << 1 | 1];
    if(rsum[rt] == r - mid) rsum[rt] += rsum[rt << 1];
    msum[rt] = max(lsum[rt << 1 | 1] + rsum[rt << 1],
                   max(msum[rt << 1], msum[rt << 1 | 1]));
}

void build(int l, int r, int rt)
{
    int mid = l + r >> 1;
    msum[rt] = lsum[rt] = rsum[rt] = r - l + 1;
    col[rt] = -1;
    if(l == r) return;
    build(lson);
    build(rson);
}

void update(int L, int R, int c, int l, int r, int rt)
{
    int mid = l + r >> 1;
    if(L <= l && r <= R)
    {
        msum[rt] = lsum[rt] = rsum[rt] = c ? 0 : r - l + 1;
        col[rt] = c;
        return;
    }
    PushDown(l, r, rt);
    if(L <= mid) update(L, R, c, lson);
    if(R > mid) update(L, R, c, rson);
    PushUp(l, r, rt);
}

int query(int w, int l, int r, int rt)
{
    int mid = l + r >> 1;
    if(l == r) return l;
    PushDown(l, r, rt);
    if(msum[rt << 1] >= w) return query(w, lson);
    else if(lsum[rt << 1 | 1] + rsum[rt << 1] >= w)
        return mid - rsum[rt << 1] + 1;
    return query(w, rson);
}

int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m) == 2)
    {
        build(1, n, 1);
        while(m --)
        {
            int op, a, b;
            scanf("%d", &op);
            if(1 == op)
            {
                scanf("%d", &a);
                if(a > msum[1]) puts("0");
                else
                {
                    int p = query(a, 1, n, 1);
                    printf("%d\n", p);
                    update(p, p + a - 1, 1, 1, n, 1);
                }
            }
            else
            {
                scanf("%d%d", &a, &b);
                update(a, a + b - 1, 0, 1, n, 1);
            }
        }
    }
    return 0;
}

 

 

 

posted on 2012-08-22 16:02  找回失去的  阅读(165)  评论(0)    收藏  举报