HDU 1166 敌兵布阵(线段树 or 二叉索引树)

时间:2023-11-26 21:10:08

http://acm.hdu.edu.cn/showproblem.php?pid=1166

题意:
第一行一个整数T,表示有T组数据。 
每组数据第一行一个正整数N(N<=50000),表示敌人有N个工兵营地,接下来有N个正整数,第i个正整数ai代表第i个工兵营地里开始时有ai个人(1<=ai<=50)。 
接下来每行有一条命令,命令有4种形式: 
(1) Add i j,i和j为正整数,表示第i个营地增加j个人(j不超过30) 
(2)Sub i j ,i和j为正整数,表示第i个营地减少j个人(j不超过30); 
(3)Query i j ,i和j为正整数,i<=j,表示询问第i到第j个营地的总人数; 
(4)End 表示结束,这条命令在每组数据最后出现;

思路:

这道题目用线段树和二叉索引树都是可以做的,给出两种做法。

 #include<iostream>
#include<cstring>
#include<algorithm>
using namespace std; const int maxn = << ; int n; struct node
{
int l, r;
int num;
}tree[maxn]; char s[];
int ans; void build_tree(int l,int r,int k)
{
tree[k].l = l;
tree[k].r = r;
tree[k].num = ; if(l==r) return; int mid = (l + r) / ;
build_tree(l, mid, * k);
build_tree(mid + , r, * k + );
} void insert(int x, int i, int k)
{
if (tree[k].l == tree[k].r && tree[k].l == i)
{
tree[k].num += x;
return;
}
int mid = (tree[k].l + tree[k].r) / ;
if (i <= mid) insert(x, i, * k);
else insert(x, i, * k + );
tree[k].num = tree[ * k].num + tree[ * k + ].num;
} void search(int l,int r,int k)
{
if (tree[k].l == l && tree[k].r == r)
{
ans += tree[k].num;
return;
}
int mid = (tree[k].l + tree[k].r) / ;
if (r <= mid) search(l, r, * k);
else if(l > mid) search(l, r, * k + );
else
{
search(l, mid, * k);
search(mid + , r, * k + );
}
} int main()
{
//freopen("D:\\txt.txt", "r", stdin);
int T;
scanf("%d", &T);
int x, y;
int kase = ;
while (T--)
{
scanf("%d", &n);
build_tree(, n, );
int a;
for (int i = ; i <= n; i++)
{
cin >> a;
insert(a, i, );
}
printf("Case %d:\n", ++kase);
while (scanf("%s",&s) && s[] != 'E')
{
scanf("%d%d", &x, &y);
if (s[] == 'Q')
{
ans = ;
search(x, y, );
cout << ans << endl;
}
else if (s[] == 'A')
{
insert(y, x, );
}
else
{
insert(-y, x, );
}
}
}
}
 #include<iostream>
#include<cstring>
#include<algorithm>
using namespace std; const int maxn = + ; int c[maxn];
int n;
char s[]; int lowbit(int x)
{
return x&-x;
} int sum(int x)
{
int num = ;
while (x > )
{
num += c[x];
x -= lowbit(x);
}
return num;
} void add(int x, int d)
{
while (x <= n)
{
c[x] += d;
x += lowbit(x);
}
} int main()
{
//freopen("D:\\txt.txt", "r", stdin);
int T;
scanf("%d", &T);
int x, y;
int kase = ;
while (T--)
{
memset(c, , sizeof(c));
cin >> n;
int a;
for (int i = ; i <= n; i++)
{
cin >> a;
add(i, a);
}
printf("Case %d:\n", ++kase);
while (scanf("%s",&s) && s[] != 'E')
{
scanf("%d%d", &x, &y);
if (s[] == 'Q')
{
cout << sum(y) - sum(x - ) << endl;
}
else if (s[] == 'A')
{
add(x, y);
}
else
{
add(x, -y);
}
}
}
}