本文主要是介绍POJ 3468 A Simple Problem with Integers （线段树区间更新模板），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！
POJ 3468
老是忘。。
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
typedef long long ll;
struct node {int l, r, m;ll val, mark; //注意此处标记和值都要定义为long long
};
const int MAXN = 100100;
node T[MAXN * 4 + 10];
ll a[MAXN];
void build(int root, int begin, int end) {T[root].mark = 0;T[root].l = begin;T[root].r = end;T[root].m = (begin + end) >> 1;if(begin == end) {T[root].val = a[begin];}else {build(root << 1, begin, T[root].m);build(root << 1 | 1, T[root].m + 1, end);T[root].val = T[root << 1].val + T[root << 1 | 1].val;}
}
ll query(int root, int l, int r) {if(T[root].l == l && T[root].r == r) {return T[root].val + T[root].mark * (r - l + 1);}else {if(T[root].mark != 0) {T[root << 1].mark += T[root].mark;T[root << 1 | 1].mark += T[root].mark;T[root].val += T[root].mark * (T[root].r - T[root].l + 1);T[root].mark = 0;}if(r <= T[root].m) return query(root << 1, l, r);else if(l > T[root].m) return query(root << 1 | 1, l, r);else return query(root << 1, l, T[root].m) + query(root << 1 | 1, T[root].m + 1, r);}
}
void updata(int root, int l, int r, int add) {if(T[root].l == l && T[root].r == r) {T[root].mark += add;}else {T[root].val += add * (r - l + 1);if(r <= T[root].m) updata(root << 1, l, r, add);else if(l > T[root].m) updata(root << 1 | 1, l, r, add);else {updata(root << 1, l, T[root].m, add);updata(root << 1 | 1, T[root].m + 1, r, add);}}
}
int main() {int n, q;while(~scanf("%d %d", &n, &q)) {int i;for(i = 1; i <= n; i++) {scanf("%I64d", a + i);}getchar();build(1, 1, n);char ch;int l, r, add;for(i = 0; i < q; i++) {scanf("%c %d %d", &ch, &l, &r);getchar();if(ch == 'Q') {printf("%I64d\n", query(1, l, r));}if(ch == 'C') {scanf("%d", &add);getchar();updata(1, l, r, add);}}}return 0;
}
另附一个线段树模板：
#include<stdio.h>
#include<algorithm>
#include<string.h>
#define ll __int64
#define M 100007
using namespace std;
struct node
{ll l,r,mid,val,mark;
}tree[M<<2];
ll s[M];
void build(ll left,ll right,ll i)//建树
{tree[i].l=left;tree[i].r=right;tree[i].mid=(left+right)>>1;tree[i].mark=0;if(left==right){tree[i].val=s[left]; return;}build(left,tree[i].mid,i*2);build(tree[i].mid+1,right,i*2+1);tree[i].val=tree[i*2].val+tree[i*2+1].val;
}
void update(int pos,ll val,int i)//点更新
{tree[i].val+=val;if(tree[i].l==tree[i].r)return;if(pos<=tree[i].mid)update(pos,val,i*2);else update(pos,val,i*2+1);
}
//ll query(int left,int right,int i)//点查询
//{
// if(tree[i].l==left&&tree[i].r==right)return tree[i].val;
// if(right<=tree[i].mid)query(left,right,i*2);
// else if(left>tree[i].mid)query(left,right,i*2+1);
// else return query(left,tree[i].mid,i*2)+query(tree[i].mid+1,right,i*2+1);
//}
void update(int left,int right,ll val,int i)//区间更新
{if(tree[i].l==left&&tree[i].r==right){tree[i].mark+=val;return;}tree[i].val+=val*(right-left+1);if(tree[i].mid<left)update(left,right,val,2*i+1);else if(tree[i].mid>=right)update(left,right,val,2*i);else{update(left,tree[i].mid,val,2*i);update(tree[i].mid+1,right,val,2*i+1);}
}
ll query(int left,int right,int i)//区间查询
{if(tree[i].l==left&&tree[i].r==right) return tree[i].val+tree[i].mark*(right-left+1);if(tree[i].mark!=0){tree[i*2].mark+=tree[i].mark;tree[i*2+1].mark+=tree[i].mark;tree[i].val+=(tree[i].r-tree[i].l+1)*tree[i].mark;tree[i].mark=0;}if(tree[i].mid>=right){return query(left,right,i*2);}else if(tree[i].mid<left){return query(left,right,i*2+1);}else{return query(left,tree[i].mid,i*2)+query(tree[i].mid+1,right,i*2+1);}
}
int main()
{ll n,m,i,j,k;ll l,f,num;char str[5];while(scanf("%I64d%I64d",&n,&m)!=EOF){for(int i=1;i<=n;i++)scanf("%I64d",&s[i]);build(0,n,1);for(i=0;i<m;i++){// printf("i=%I64d",i);scanf("%s",&str);if(str[0]=='Q'){scanf("%I64d%I64d",&l,&f);printf("%I64d\n",query(l,f,1));}if(str[0]=='C'){scanf("%I64d%I64d%I64d",&l,&f,&num);update(l,f,num,1);}}}return 0;
}这篇关于POJ 3468 A Simple Problem with Integers （线段树区间更新模板）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
