hdu 1166 敌兵布阵（单点更新）

hdu 1166 敌兵布阵（基本操作）

有三种操作：询问区间总和，增加某个兵营的兵的数目，减少某个兵营的兵的数目。实际上也只有两个。

在更新的时候，每到一个区间就把当前区间的sum增加对应的数目，到达叶子结点是返回。这样就可以不会回溯去更新父亲结点的值。查询的时候，如果区间完全匹配，直接返回区间的sum值，否则向下寻找，直到完全匹配，然后返回它们的和就可以。这时候，结果里保存的边界是它们真正的边界。

/*更新了代码风格后*/

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define MID(a,b) (a+((b-a)>>1))
const int N=50005;

struct node
{
	int lft,rht;
	int sum;
	int mid(){return MID(lft,rht);}
};

int y[N],n;

struct Segtree
{
	node tree[N*4];
	void build(int lft,int rht,int ind)
	{
		tree[ind].lft=lft;	tree[ind].rht=rht;
		tree[ind].sum=0;
		if(lft==rht) tree[ind].sum=y[lft];
		else 
		{
			int mid=tree[ind].mid();
			build(lft,mid,LL(ind));
			build(mid+1,rht,RR(ind));
			tree[ind].sum=tree[LL(ind)].sum+tree[RR(ind)].sum;
		}
	}
	void updata(int pos,int ind,int valu)
	{
		if(tree[ind].lft==tree[ind].rht) tree[ind].sum+=valu;
		else 
		{
			int mid=tree[ind].mid();
			if(pos<=mid) updata(pos,LL(ind),valu);
			else updata(pos,RR(ind),valu);
			tree[ind].sum=tree[LL(ind)].sum+tree[RR(ind)].sum;
		}
	}
	int query(int st,int ed,int ind)
	{
		int lft=tree[ind].lft,rht=tree[ind].rht;
		if(st<=lft&&rht<=ed) return tree[ind].sum;
		else 
		{
			int mid=tree[ind].mid();
			int sum1=0,sum2=0;
			if(st<=mid) sum1=query(st,ed,LL(ind));
			if(ed>mid) sum2=query(st,ed,RR(ind));
			return sum1+sum2;
		}
	}
}seg;
int main()
{
	int t,t_cnt=0;
	scanf("%d",&t);
	while(t--)
	{
		int a,b;
		char str[10];
		scanf("%d",&n);
		for(int i=1;i<=n;i++) scanf("%d",&y[i]);
		seg.build(1,n,1);

		printf("Case %d:\n",++t_cnt);
		while(1)
		{
			scanf("%s",str);
			if(strcmp(str,"End")==0) break;

			scanf("%d%d",&a,&b);
			if(strcmp(str,"Add")==0) seg.updata(a,1,b);
			else if(strcmp(str,"Sub")==0) seg.updata(a,1,-b);
			else printf("%d\n",seg.query(a,b,1));
		}
	}
	return 0;
}

/*代码风格更新前*/

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1000005;
int a[N];
struct node
{
    int left,right,sum;
    int mid(){ return left+(right-left)/2;}
};
struct Segtree
{
    node tree[N*4];
    void build(int left,int right,int p)
    {
        tree[p].left=left;
        tree[p].right=right;
        tree[p].sum=0;
        if(left<right)
        {
            int mid=tree[p].mid();
            build(left,mid,p*2);
            build(mid+1,right,p*2+1);
        }
        else if(left==right)
        {
            tree[p].sum=a[left];
            return;
        }
        tree[p].sum=tree[p*2].sum+tree[p*2+1].sum;
    }
    void updata(int pos,int p,int data)
    {
        tree[p].sum+=data;
        int mid=tree[p].mid();
        if(tree[p].left==pos&&tree[p].right==pos) return;
        else if(pos<=mid) updata(pos,p*2,data);
        else if(pos>mid) updata(pos,p*2+1,data);
    }
    int query(int be,int end,int p)
    {
        int left=tree[p].left;
        int right=tree[p].right;
        if(be<=left&&right<=end)
        {
            return tree[p].sum;
        }
        else
        {
            int sum1=0,sum2=0;
            int mid=tree[p].mid();
            if(be<=mid) sum1=query(be,end,p*2);
            if(end>mid) sum2=query(be,end,p*2+1);
            return sum1+sum2;
        }
    }
}seg;
int main()
{
    int t,t_cnt=0;
    scanf("%d",&t);
    while(t--)
    {
        int n,x,y;
        char temp[10];
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);

        printf("Case %d:\n",++t_cnt);

        seg.build(1,n,1);
        while(scanf("%s",temp)!=EOF)
        {
            if(strcmp(temp,"End")==0) break;
            else if(strcmp(temp,"Add")==0)
            {
                scanf("%d%d",&x,&y);
                seg.updata(x,1,y);
            }
            else if(strcmp(temp,"Sub")==0)
            {
                scanf("%d%d",&x,&y);
                seg.updata(x,1,-y);
            }
            else if(strcmp(temp,"Query")==0)
            {
                scanf("%d%d",&x,&y);
                printf("%d\n",seg.query(x,y,1));
            }
        }
    }
    return 0;
}