HDOJ 1166 敌兵布阵 （线段树）

#include <stdio.h>
typedef struct
{
int l, r, count;
}node;
node tree[4*50005];
void Init(int root, int left, int right)
{
int m = (left+right)/2;
if(left == right)//叶节点为单个的数
{
tree[root].l = tree[root].r = left;
scanf("%d",&tree[root].count);
return ;
}
tree[root].l = left; tree[root].r = right;
Init(2*root, left, m);
Init(2*root+1, m+1, right);
tree[root].count = tree[2*root].count+tree[2*root+1].count;
}
void modify(int root, int a, int b)//a为第几个营地， b为要修改的数
{
int m = (tree[root].l+tree[root].r)/2;
if(tree[root].l == tree[root].r && tree[root].l == a)//找到叶节点
{
tree[root].count += b;
return ;
}
if(a>m)
modify(2*root+1, a, b);
else
modify(2*root, a, b);
tree[root].count = tree[2*root].count+tree[2*root+1].count;
}
int search(int root, int left, int right)
{
int l = tree[root].l ,r = tree[root].r, m = (tree[root].l+tree[root].r)/2, sum = 0;
if(left<=l && right>=r)//当前区间属于查询区间
return  sum += tree[root].count;
if(left>m)//查询区间属于树的右区间
return sum = search(2*root+1, left, right);
else if(right<=m)//查询区见属于树的左区间
return sum = search(2*root, left, right);
else
return sum = search(2*root+1, left, right)+search(2*root, left, right);
}
int main()
{
int t = 0, n = 0, count = 0, a = 0, b = 0;
char ch[10];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
Init(1, 1, n);
printf("Case %d:\n", ++count);
while(1)
{
scanf("%s",ch);
if(ch[0] == 'E')
break;
if(ch[0] == 'A')
{
scanf("%d %d",&a,&b);
modify(1,a,b);
}
else if(ch[0] == 'S')
{
scanf("%d %d",&a,&b);
modify(1,a,-b);
}
else
{
scanf("%d %d",&a,&b);
printf("%d\n",search(1,a,b));
}
}
}
return 0;
}