hdu 3954(区间更新)

来源：互联网发布：mysql 更新字段加一编辑：程序博客网时间：2024/05/21 09:26

题意：有n个人，每个人初始等级是1，初始经验值是0，一共有K个等级，给出到达每个等级需要的经验值，然后有q个操作，W l r e把第l个人到第r个人的经验值都加上其对应的等级×e，Q l r询问l到r人中经验值的最大值。
题解：刚开始用单点修改把每个人经验值改掉，毫无疑问的超时了，然后就想怎么把每个人获得不同的经验值在区间表示，苦逼的想不出来，然后看了题解。。。因为要维护区间最大经验值，所以就维护一个最小到下一等级需要的经验值need，need[k] = (下一等级经验值-当前经验值) / 当前等级 + 余数，这样消除不同等级的人所加经验值的不同。

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;const int N = 10005;int level[N << 2], n, K, q;ll Exp[N << 2], need[N << 2], flag[N << 2], ex[15];void pushup(int k) {    Exp[k] = max(Exp[k * 2], Exp[k * 2 + 1]);    level[k] = max(level[k * 2], level[k * 2 + 1]);    need[k] = min(need[k * 2], need[k * 2 + 1]);}void pushdown(int k) {    if (flag[k]) {        Exp[k * 2] += flag[k] * level[k * 2];        need[k * 2] -= flag[k];        flag[k * 2] += flag[k];        Exp[k * 2 + 1] += flag[k] * level[k * 2 + 1];        need[k * 2 + 1] -= flag[k];        flag[k * 2 + 1] += flag[k];        flag[k] = 0;    }}void build(int k, int left, int right) {    need[k] = ex[2];    Exp[k] = flag[k] = 0;    level[k] = 1;    if (left != right) {        int mid = (left + right) / 2;        build(k * 2, left, mid);        build(k * 2 + 1, mid + 1, right);    }}void modify(int k, int left, int right, int l, int r, ll v) {    if (l <= left && right <= r) {        if (v >= need[k]) {            if (left == right) {                Exp[k] += level[k] * v;                while (Exp[k] >= ex[level[k] + 1]) level[k]++;                ll temp = ex[level[k] + 1] - Exp[k];                need[k] = temp / level[k] + (temp % level[k] != 0);            }            else {                pushdown(k);                int mid = (left + right) / 2;                if (l <= mid) modify(k * 2, left, mid, l, r, v);                if (r > mid) modify(k * 2 + 1, mid + 1, right, l, r, v);                pushup(k);            }        }        else {            Exp[k] += v * level[k];            need[k] -= v;            flag[k] += v;        }        return;    }    pushdown(k);    int mid = (left + right) / 2;    if (l <= mid) modify(k * 2, left, mid, l, r, v);    if (r > mid) modify(k * 2 + 1, mid + 1, right, l, r, v);    pushup(k);    return;}ll query(int k, int left, int right, int l, int r) {    if (l <= left && right <= r)        return Exp[k];    pushdown(k);    int mid = (left + right) / 2;    ll res = 0;    if (l <= mid)        res = max(res, query(k * 2, left, mid, l, r));      if (r > mid)        res = max(res, query(k * 2 + 1, mid + 1, right, l, r));    pushup(k);    return res;}int main() {    int t, cas = 1;    scanf("%d", &t);    ex[1] = 0;    while (t--) {        scanf("%d%d%d", &n, &K, &q);        for (int i = 2; i <= K; i++)            scanf("%lld", &ex[i]);        ex[K + 1] = (ll)(1 << 30);        build(1, 1, n);        char op[5];        int l, r;        ll e;        printf("Case %d:\n", cas++);        while (q--) {            scanf("%s%d%d", op, &l, &r);            if (op[0] == 'W') {                scanf("%lld", &e);                modify(1, 1, n, l, r, e);            }            else {                printf("%lld\n", query(1, 1, n, l, r));            }        }        printf("\n");    }    return 0;}

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