HDU 3954 区间更新区间查询 打怪升级

题意：

T个测试数据

n个英雄 满级K级(初始化英雄都为1级0经验) que个操作

K-1 表示升级所需经验

W u v exp 表示 [u, v] 区间的英雄获得打怪经验为 exp

Q u v 询问区间经验值最高的 经验值大小



获得的经验为 当前等级*打怪经验

#include<string.h>
#include<stdio.h>
#include<queue>
#include<math.h>

#define inf 33554432
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define Mid(x,y) ((x+y)>>1)
#define ll int
#define N 10010
using namespace std;

inline int Max(int a,int b){return a<b?b:a;}
inline int Min(int a,int b){return a>b?b:a;}

struct node{
	int l, r;
	int grade, ex;//grade 为区间中最高等级 ex为区间中最大的 英雄的总经验(非经验条经验) 
	int lazy;//表示打怪获得经验
	int need;//区间中升级所需最少经验的英雄 所需的打怪经验
}tree[N*8];

int expe[12], K;
void updata_grade(int id){
	for(int i = tree[id].grade; i < K;i++)
		if(tree[id].ex >= expe[i])
			tree[id].grade = i + 1;
}

void updata_down(int id){
	tree[L(id)].ex += tree[L(id)].grade*tree[id].lazy;
	tree[L(id)].lazy += tree[id].lazy;
	tree[L(id)].need -= tree[id].lazy;

	tree[R(id)].ex += tree[R(id)].grade*tree[id].lazy;
	tree[R(id)].lazy += tree[id].lazy;
	tree[R(id)].need -= tree[id].lazy;

	tree[id].lazy = 0;
}
void updata_up(int id){
	tree[id].ex = Max(tree[L(id)].ex, tree[R(id)].ex);
	tree[id].grade = Max(tree[L(id)].grade, tree[R(id)].grade);
	tree[id].need = Min(tree[L(id)].need, tree[R(id)].need);

}
void build(int l, int r, int id){
	tree[id].l = l , tree[id].r = r;
	tree[id].ex = 0;
	tree[id].grade = 1;
	tree[id].lazy = 0;
	tree[id].need = expe[1];

	if(l == r)return ;
	int mid = Mid(l,r);
	build(l,   mid, L(id));
	build(mid+1, r, R(id));
}
void updata(int l, int r, int id, int Exp){
	updata_down(id);
	if(tree[id].l == tree[id].r)
		{
			tree[id].ex += Exp* tree[id].grade;
			updata_grade(id);
			tree[id].need =(expe[tree[id].grade]-tree[id].ex)/tree[id].grade+((expe[tree[id].grade]-tree[id].ex)%tree[id].grade!=0);  
			return;
		}
	if(l == tree[id].l && tree[id].r == r)
	{
		if(tree[id].need > Exp) //获得打怪经验后区间内英雄都不会升级
		{
			tree[id].lazy += Exp;
			tree[id].ex += tree[id].grade*Exp;
			tree[id].need -= Exp;
			return ;
		}
		updata_down(id);//有英雄要升级了
		updata(tree[L(id)].l, tree[L(id)].r, L(id), Exp);
		updata(tree[R(id)].l, tree[R(id)].r, R(id), Exp);
		updata_up(id);
		return ;
	}

	int mid = Mid(tree[id].l, tree[id].r);
	if(r<=mid) updata(l, r, L(id), Exp);
	else if(l>mid)updata(l, r, R(id), Exp);
	else
	{
		updata(l, mid, L(id), Exp);
		updata(mid+1, r, R(id), Exp);
	}
	updata_up(id); ///
}
int query(int l, int r, int id){
	if(l == tree[id].l && tree[id].r == r)
	{
		return tree[id].ex;
	}

	updata_down(id);
	int mid = Mid(tree[id].l, tree[id].r);
	if(r<=mid) return query(l, r, L(id));
	else if(mid<l) return query(l, r, R(id));
	else 
		return Max( query(l, mid, L(id)), query(mid+1, r, R(id)));
}

int main(){
	expe[0] = 0;
	int T, Cas = 1;scanf("%d",&T);
	int n, que, i, u, v, EXP;

	while(T--){
		scanf("%d %d %d",&n,&K,&que);
		for(i = 1; i < K; i++)scanf("%d",&expe[i]); 
		expe[K] = inf;

		build(1, n, 1);
		printf("Case %d:\n",Cas++);
		while(que--)
		{
			char c = 'h'; while(c!='W' && c!='Q')c = getchar();
			scanf("%d %d",&u,&v);
			if(c == 'W')
			{ scanf("%d",&EXP); updata(u, v, 1, EXP); }
			else 
				printf("%d\n",query(u, v, 1));
		}
		printf("\n");
	}
	return 0;
}