hdu 2871 Memory Control(线段树)

题目链接：hdu 2871 Memory Control

题目大意：模拟一个内存分配机制。

• Reset：重置，释放全部空间
• New x：申请内存为x的空间，输出左地址
• Free x：释放地址x所在的内存块
• Get x：查询第x个内存块，输出左地址

解题思路：一開始全用线段树去做，写的乱七八糟，事实上仅仅要用线段树维护可用内存。然后用户一个vector记录全部的内存块。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
const int maxn = 50005;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], set[maxn << 2];
int L[maxn << 2], R[maxn << 2], S[maxn << 2];

inline int length (int u) {
    return rc[u] - lc[u] + 1;
}

inline void maintain (int u, int v) {
    set[u] = v;
    L[u] = R[u] = S[u] = (v ? 0 : length(u));
}

inline void pushup (int u) {
    S[u] = max( max(S[lson(u)], S[rson(u)]), L[rson(u)] + R[lson(u)]);
    L[u] = L[lson(u)] + (L[lson(u)] == length(lson(u)) ? L[rson(u)] : 0);
    R[u] = R[rson(u)] + (R[rson(u)] == length(rson(u)) ? R[lson(u)] : 0);
}

inline void pushdown (int u) {
    if (set[u] != -1) {
        maintain(lson(u), set[u]);
        maintain(rson(u), set[u]);
        set[u] = -1;
    }
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    set[u] = -1;

    if (l == r) {
        maintain(u, 0);
        return;
    }

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int l, int r, int v) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, v);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, v);
    if (r > mid)
        modify(rson(u), l, r, v);
    pushup(u);
}

int query (int u, int len) {
    if (S[u] < len)
        return 0;

    if (lc[u] == rc[u])
        return lc[u];

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2, ret;
    if (S[lson(u)] >= len)
        ret = query(lson(u), len);
    else if (L[rson(u)] + R[lson(u)] >= len)
        ret = mid - R[lson(u)] + 1;
    else
        ret = query(rson(u), len);
    pushup(u);
    return ret;
}

typedef pair<int, int> pii;
int N, M;
vector<pii> list;

int find (int k) {
    int l = 0, r = list.size() - 1;
    while (l <= r) {
        int mid = (l + r) / 2;
        if (list[mid].first > k)
            r = mid - 1;
        else
            l = mid + 1;
    }
    return l;
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        build (1, 1, N);
        list.clear();

        int k;
        char op[5];
        while (M--) {
            scanf("%s", op);
            if (op[0] == ‘R‘) {
                modify(1, 1, N, 0);
                list.clear();
                printf("Reset Now\n");
            } else {
                scanf("%d", &k);
                if (op[0] == ‘N‘) {
                    int x = query(1, k);

                    if (x) {
                        modify(1, x, x + k - 1, 1);
                        pii u = make_pair(x, x + k - 1);
                        list.insert(list.begin() + find(x), u);
                        printf("New at %d\n", x);
                    } else
                        printf("Reject New\n");

                } else if (op[0] == ‘F‘) {
                    int x = find(k) - 1;

                    if (x != -1 && k <= list[x].second) {
                        modify(1, list[x].first, list[x].second, 0);
                        printf("Free from %d to %d\n", list[x].first, list[x].second);
                        list.erase(list.begin() + x);
                    } else
                        printf("Reject Free\n");

                } else if (op[0] == ‘G‘) {
                    if (k <= list.size()) {
                        printf("Get at %d\n", list[k-1].first);
                    } else
                        printf("Reject Get\n");
                }
            }
        }
        printf("\n");
    }
    return 0;
}