hdu 3954 Level up(线段树)

题目链接：hdu 3954 Level up

题目大意：N个英雄，M个等级，初始等级为1，给定每一个等级须要的经验值，Q次操作，操作分两种，W l r x：表示l~r之间的英雄每一个人杀了x个怪物；Q l r：表示询问l~r之间经验值最大的英雄经验值为多少。

每轮杀怪，每仅仅怪物的经验和当前等级成正比。

解题思路：线段树维护，每一个节点维护最大值，区间内还须要杀多少怪就能升级的最小值，假设这个最小值为0。就要将懒惰标记pushdown到最底层，将英雄升级。

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

using namespace std;

const int maxn = 10005;
const int INF = 0x3f3f3f3f;

int N, K, Q, need[15];

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2];
int ad[maxn << 2], nd[maxn << 2], rk[maxn << 2];

void maintain(int u, int d);

inline void pushup(int u) {
    s[u] = max(s[lson(u)], s[rson(u)]);
    rk[u] = max(rk[lson(u)], rk[rson(u)]);
    nd[u] = min(nd[lson(u)], nd[rson(u)]);
}

inline void pushdown(int u) {
    if (ad[u]) {
        maintain(lson(u), ad[u]);
        maintain(rson(u), ad[u]);
        ad[u] = 0;
    }
}

inline void levelup(int u) {
    ad[u] = 0;
    while (s[u] >= need[rk[u] + 1])
        rk[u]++;
    int tmp = need[rk[u] + 1] - s[u];
    nd[u] = tmp / rk[u] + (tmp % rk[u] ? 1 : 0);
}

void maintain (int u, int d) {
    nd[u] -= d;
    ad[u] += d;
    s[u] += rk[u] * d;

    if (nd[u] <= 0) {
        if (lc[u] == rc[u])
            levelup(u);
        else {
            pushdown(u);
            pushup(u);
        }
    }
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    s[u] = ad[u] = nd[u] = rk[u] = 0;

    if (l == r) {
        rk[u] = 1;
        nd[u] = need[2];
        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 d) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, d);
        return;
    }

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

int query (int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return s[u];

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2, ret = 0;
    if (l <= mid)
        ret = max(ret, query(lson(u), l, r));
    if (r > mid)
        ret = max(ret, query(rson(u), l, r));
    pushup(u);
    return ret;
}

void init () {
    scanf("%d%d%d", &N, &K, &Q);

    for (int i = 2; i <= K; i++)
        scanf("%d", &need[i]);
    need[K+1] = INF;
    build(1, 1, N);
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();
        printf("Case %d:\n", kcas);

        char op[5];
        int t, l, r;
        while (Q--) {
            scanf("%s%d%d", op, &l, &r);
            if (op[0] == 'W') {
                scanf("%d", &t);
                modify(1, l, r, t);
            } else
                printf("%d\n", query(1, l, r));
        }
        printf("\n");
    }
    return 0;
}







本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5233571.html，如需转载请自行联系原作者