HDU1754 && HDU1166 线段树模板题

时间:2023-03-09 05:38:10
HDU1754 && HDU1166 线段树模板题

HDU1754

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1754

题目分析:对于给出的一个很长的区间,对其进行单点更新值和区间求最大值的操作,由于查询的区间很大,且查询次数多,这里用线段树求解将是十分合适的

注意点:1.对于存放线段树的数组大小需要开大一些

    2.对于c语言的字符输入%c之前需要加一个空格保证输入准确

 #include<iostream>
#include<string.h>
using namespace std; const int N = ;
int grade[N];
int tree[N<<]; //这里建立的树的数组大小需要是N的4倍 否则不够用
int n, m; int max(int a, int b){
if(a > b) return a;
else return b;
} void build_tree(int start, int end, int node){ //线段树的建立
if(start == end){
tree[node] = grade[start];
}else{
int mid = (start + end) / ;
int left_node = node*;
int right_node = node*+; build_tree(start, mid, left_node);
build_tree(mid+, end, right_node);
tree[node] = max(tree[left_node], tree[right_node]);
}
} void update_tree(int start, int end, int node, int index, int value){ //单点更新值
if(start == end){
tree[node] = value;
}else{
int mid = (start + end) / ;
int left_node = node*;
int right_node = node*+; if(index <= mid)
update_tree(start, mid, left_node, index, value);
else
update_tree(mid+, end, right_node, index, value);
tree[node] = max(tree[left_node], tree[right_node]);
}
} int search_tree(int start, int end, int node, int l, int r){ //区间查询最大值
if(l > end || r < start){
return ;
}else if(l <= start && r >= end){
return tree[node];
}else if(start == end){      //这里的个递归出口放在后面是有原因的,有可能存在start==end 但是l和r根本和start end没有交集的情况,所以先判去了后者
return tree[node];
}
int mid = (start + end) / ;
int left_node = node*;
int right_node = node*+; int left_max = search_tree(start, mid, left_node, l, r);
int right_max = search_tree(mid+, end, right_node, l, r);
int ans = max(left_max, right_max);
return ans;
} int main(){
while(scanf("%d%d", &n, &m) != EOF){
for(int i = ; i <= n; i++)
scanf("%d", &grade[i]);
build_tree(, n, );
for(int i = ; i <= m; i++){
char c;
int a, b;
scanf(" %c %d %d", &c, &a, &b); //对于c语言的输入字符在%c之前需要一个空格,否则c就读取不到需要的字符了
if(c == 'U') update_tree(, n, , a, b);
else{
int ans = search_tree(, n, , a, b);
printf("%d\n", ans);
}
}
}
return ;
}

HDU1166

题目分析:

也是一题线段树的模板题,单点更新和区间查询求和(本质上和区间求最大值是一样的)

代码:

 #include<iostream>
#include<string>
#include<string.h>
using namespace std; const int N = ;
int peo[N];
int tree[N<<]; void build_tree(int start, int end, int node){
if(start == end){
tree[node] = peo[start];
}else{
int mid = (start + end) / ;
int left_node = node*;
int right_node = node*+; build_tree(start, mid, left_node);
build_tree(mid+, end, right_node);
tree[node] = tree[left_node] + tree[right_node];
}
} void update_tree(int start, int end, int node, int index, int value){
if(start == end){
tree[node] += value;
}else{
int mid = (start + end) / ;
int left_node = node*;
int right_node = node*+; if(index <= mid)
update_tree(start, mid, left_node, index, value);
else
update_tree(mid+, end, right_node, index, value);
tree[node] = tree[left_node] + tree[right_node];
}
} int query_tree(int start, int end, int node, int l, int r){
if(end < l || start > r){
return ;
}else if(start >= l && end <= r){
return tree[node];
}else if(start == end){
return tree[node];
}
int mid = (start + end) / ;
int left_node = node*;
int right_node = node*+; int left_sum = query_tree(start, mid, left_node, l, r);
int right_sum = query_tree(mid+, end, right_node, l, r);
int ans = left_sum + right_sum;
return ans;
} int main(){
int t;
scanf("%d", &t);
int cnt = ;
while(t--){
printf("Case %d:\n", cnt++);
int n;
scanf("%d", &n);
for(int i = ; i <= n; i++)
scanf("%d", &peo[i]);
build_tree(, n, );
string s;
int a, b;
while(cin>>s){
if(s == "End") break;
scanf("%d%d", &a, &b);
if(s == "Query"){
int ans = query_tree(, n, , a, b);
printf("%d\n", ans);
}
if(s == "Add"){
update_tree(, n, , a, b);
}
if(s == "Sub"){
update_tree(, n, , a, -b);
}
}
}
return ;
}