hdu 1166 敌兵布阵

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思路: 线段树单点更新

分析:

1 题目给定n个兵营的人数，现在有三种操作

(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;

2 最简单的线段树单点更新的题目

代码:

/************************************************
 * By: chenguolin                               * 
 * Date: 2013-09-01                             *
 * Address: http://blog.csdn.net/chenguolinblog *
 ************************************************/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

#define Lson(x) (x<<1)
#define Rson(x) (Lson(x)|1)
#define Mid(x,y) ((x+y)>>1)
#define Sum(x,y) (x+y)

const int N = 10;
const int MAXN = 50010;

int n;
int num[MAXN];
struct Node{
    int left;
    int right;
    int sum;
};
Node node[4*MAXN];

void push_up(int pos){
    node[pos].sum = Sum(node[Lson(pos)].sum,node[Rson(pos)].sum);
}

void init(int left , int right , int pos){
    node[pos].left = left;
    node[pos].right = right;
    if(left == right){
       node[pos].sum = num[left];
       return;
    }
    int x = Mid(left , right);
    init(left , x , Lson(pos));
    init(x+1 , right , Rson(pos));
    push_up(pos);
}

void update(int index , int val , int pos){
    if(node[pos].left == node[pos].right){
        node[pos].sum += val;
        return;
    }
    int x = Mid(node[pos].left , node[pos].right);
    if(index <= x)
        update(index , val , Lson(pos));
    else
        update(index , val , Rson(pos));
    push_up(pos);
}

int query(int left , int right , int pos){
    if(node[pos].left == left && node[pos].right == right)
        return node[pos].sum;
    int x = Mid(node[pos].left , node[pos].right);
    if(right <= x)
        return query(left , right , Lson(pos));
    else if(left > x)
        return query(left , right , Rson(pos));
    else
        return query(left , x , Lson(pos))+query(x+1 , right , Rson(pos));
}


void input(){
    char str[N];
    scanf("%d" , &n);
    for(int i = 1 ; i <= n ; i++)
        scanf("%d" , &num[i]);
    init(1 , n , 1);
    getchar();
    int x , y;
    while(scanf("%s" , str) && str[0] != 'E'){
        scanf("%d%d%*c" , &x , &y);
        if(str[0] == 'Q')
            printf("%d\n" , query(x , y , 1));
        else if(str[0] == 'A')
            update(x , y , 1);
        else
            update(x , -y , 1);
    }
}

int main(){
    int Case;
    int cas = 1;
    scanf("%d" , &Case);
    while(Case--){
        printf("Case %d:\n" , cas++);
        input();
    }
    return 0;
}

思路：树状数组
分析：
1 对于区间求和的问题一般利用树状数组比线段树来的方便
2 树状数组建立就是在输入数据的时候做n次的update()。

代码：

/***********************************************
* By: chenguolin                               * 
* Date: 2013-08-20                             *
* Address: http://blog.csdn.net/chenguolinblog *
***********************************************/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 50010;

int n;
int treeNum[MAXN];

int lowbit(int x){
    return x&(-x);
}

int getSum(int x){
    int sum = 0;
    while(x){
        sum += treeNum[x];
        x -= lowbit(x);
    }
    return sum;
}

void add(int x , int val){
    while(x < MAXN){
         treeNum[x] += val;
         x += lowbit(x);
    }
}

void solve(){
    int x , y;
    char str[10];
    memset(treeNum , 0 , sizeof(treeNum));
    for(int i = 1 ; i <= n ; i++){
        scanf("%d%*c" , &x);
        add(i , x);
    }
    while(scanf("%s" , str) && str[0] != 'E'){
        scanf("%d%d%*c" , &x , &y);
        if(str[0] == 'Q'){
           int ans = getSum(y); 
           ans -= getSum(x-1);
           printf("%d\n" , ans);
        } 
        else if(str[0] == 'A'){
           add(x , y); 
        } 
        else{ 
           add(x , -y); 
        } 
    }
}

int main(){
    int cas = 1;
    int Case;
    scanf("%d" , &Case);
    while(Case--){
         printf("Case %d:\n" , cas++); 
         scanf("%d" , &n);
         solve();
    }
    return 0;
}