本文主要是介绍HDU 1166 敌兵布阵(线段树树状数组)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：HDU 1166 敌兵布阵
线段树模板题同时也是树状数组模板题，两种解法：
线段树：
//Must so
#include<bits/stdc++.h>
#define mem(a,x) memset(a,x,sizeof(a))
#define sqrt(n) sqrt((double)n)
#define pow(a,b) pow((double)a,(int)b)
#define inf 1<<29
#define NN 50004
using namespace std;
const double PI = acos(-1.0);
typedef long long LL;struct Node
{int l,r,v;
}q[4*NN];
void buildtree(int i,int l,int r)
{q[i].l = l,q[i].r = r;if (l == r){scanf("%d",&q[i].v);return ;}int mid = (l+r)/2;buildtree(i<<1,l,mid);buildtree((i<<1)+1,mid+1,r);q[i].v = q[i<<1].v + q[(i<<1)+1].v;
}
void Add(int i,int a,int b)
{if (q[i].l == q[i].r &&q[i].l == a){q[i].v += b;return ;}int mid = (q[i].l+q[i].r)/2;if (a <= mid) Add(i<<1,a,b);else Add((i<<1)+1,a,b);q[i].v += b;
}
int query(int i,int l,int r)
{if (l == q[i].l&&r == q[i].r){return q[i].v;}int mid = (q[i].l+q[i].r)/2;if (r <= mid) return query(i<<1,l,r);else if (l > mid) return query((i<<1)+1,l,r);else return query(i<<1,l,mid)+query((i<<1)+1,mid+1,r);
}
int main()
{int T;scanf("%d",&T);int kas = 1;while (T--){printf("Case %d:\n",kas++);int n;scanf("%d",&n);buildtree(1,1,n);string s;while (cin>>s){if (s == "End") break;int a,b;scanf("%d%d",&a,&b);if (s == "Query"){printf("%d\n",query(1,a,b));}else if (s == "Add"){Add(1,a,b);}else{Add(1,a,-b);}}}return 0;
}
这是本渣第一道线段树的题，表示对移位运算符的优先级非常不满
树状数组：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<stdlib.h>
#include<cctype>
#include<map>
#define mem(a,x) memset(a,x,sizeof(a))
#define esp 1e-8
using namespace std;
typedef long long ll;
const int N = 50000;
int c[N+5],n;
int lowbit(int x)
{return x&(-x);
}
void Add(int i,int v)//将第i个元素的值修改成v
{while (i <= n){c[i] += v;i += lowbit(i);}
}
int Sun(int i)//求前i项和
{int s = 0;while (i > 0){s += c[i];i -= lowbit(i);}return s;
}
int main()
{int T;scanf("%d",&T);int kas = 0;while (T--){scanf("%d",&n);mem(c,0);for (int i = 1,v;i <= n;++i){scanf("%d",&v);Add(i,v);}printf("Case %d:\n",++kas);string s;while (cin>>s){if (s == "End") break;int x,y;scanf("%d %d",&x,&y);if (s == "Add"){Add(x,y);}else if (s == "Sub"){Add(x,-y);}else printf("%d\n",Sun(y)-Sun(x-1));}}return 0;
}树状数组代码更短，但是其实个人对于线段树的理解更深，不过某大神说算法竞赛优先选择用树状数组。。。
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