HDU 1166 敌兵布阵（线段树入门，单点更新）

ACM

题目地址：HDU 1166 敌兵布阵

题意： 
中文题不解释。

分析： 
经典入门题。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        1166.cpp*  Create Date: 2014-08-04 23:52:52*  Descripton:   */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)#define lson(x) ((x) << 1)#define rson(x) ((x) << 1 | 1)typedef long long ll;const int N = 50000;const int ROOT = 1;// below is sement point updatedstruct seg {ll w;};struct segment_tree { seg node[N << 2];void update(int pos) {node[pos].w = node[lson(pos)].w + node[rson(pos)].w;}void build(int l, int r, int pos) {if (l == r) {scanf("%lld", &node[pos].w);return;}int m = (l + r) >> 1;build(l, m, lson(pos));build(m + 1, r, rson(pos));update(pos);}// add the point x with yvoid modify(int l, int r, int pos, int x, ll y) {if (l == r) {node[pos].w += y;return;}int m = (l + r) >> 1;if (x <= m)modify(l, m, lson(pos), x, y);elsemodify(m + 1, r, rson(pos), x, y);update(pos);}// query the segment [x, y]ll query(int l, int r, int pos, int x, int y) {if (x <= l && r <= y)return node[pos].w;int m = (l + r) >> 1;ll res = 0;if (x <= m)res += query(l, m, lson(pos), x, y);if (y > m)res += query(m + 1, r, rson(pos), x, y);return res;}} sgm;int t, n, a;ll b;char op[10];int main() {scanf("%d", &t);repf (cas, 1, t) {printf("Case %d:\n", cas);scanf("%d", &n);sgm.build(1, n, ROOT);while (~scanf("%s", op) && op[0] != 'E') {scanf("%d%lld", &a, &b);if (op[0] == 'A') {sgm.modify(1, n, ROOT, a, b);} else if (op[0] == 'S') {sgm.modify(1, n, ROOT, a, -b);} else {printf("%lld\n", sgm.query(1, n, ROOT, a, (int)b));}}}return 0;}