HDU  1166  敌兵布阵

1
10
1 2 3 4 5 6 7 8 9 10
Query 1 3
Add 3 6
Query 2 7
Sub 10 2
Add 6 3
Query 3 10
End 

Sample Output

Case 1:
6
33
59 

题目大意：中文题目，不解释。

解题思路：与之前那题大致相同，只是之前那题要求最高分，而这题要求总和。

#include<stdio.h>
#include<string.h>
#define N 50005 * 4
int S[N], L[N], R[N], V[N];
void build(int n, int l, int r) {      //建线段树
	if (l == r) {
		R[n] = L[n] = l;
		S[n] = V[l];
	}else {
		int mid = (l + r) / 2;
		L[n * 2] = l;
		R[n * 2] = mid;
		L[n * 2 + 1] = mid + 1;
		R[n * 2 + 1] = r;
		build(n * 2 , L[n * 2], R[n * 2]);
		build(n * 2 + 1, L[n * 2 + 1], R[n * 2 + 1]);
		S[n] = S[n * 2] + S[n * 2 + 1];               //改成求和
	}
}
void modify(int n, int x, int v) {     //修改数据
	if (L[n] == x && R[n] == x) {
		S[n] += v;
		return;
	}	
	int mid = (L[n] + R[n]) / 2;
	if (x <= mid) {
		modify(n * 2, x, v);
	}
	else  {
		modify(n * 2 + 1, x, v);
	}
	S[n] = S[n * 2] + S[n * 2 + 1]; //维护线段树
}
int find(int n, int l, int r) {    //输出区间和
	if (l <= L[n] && r >= R[n]) {
		return S[n];
	}
	int mid = (L[n] + R[n]) / 2;
	int sum = 0;
	if (l <= mid) {
		sum += find(n * 2, l, r);
	}
	if (r > mid) {
		sum += find(n * 2 + 1, l, r);
	}
	return sum;
}
int main() {
	int cnt, cnt2 = 1;
	while (scanf("%d\n", &cnt) == 1) {
		while (cnt--) {
			memset(S, 0, sizeof(S));
			memset(L, 0, sizeof(L));
			memset(R, 0, sizeof(R));
			memset(V, 0, sizeof(V));
			int cnt3 = 0, n;
			printf("Case %d:\n", cnt2++);
			scanf("%d", &n);
			for (int i = 1; i <= n; i++) {
				scanf("%d", &V[i]);
			}
			L[1] = 1;
			R[1] = n;
			build(1, 1, n);
			char O[100];
			int a, b;
			memset(O, 0, sizeof(O));
			while (scanf("%s", O)) {
				if (O[0] == 'A') {
					scanf("%d %d", &a, &b);
					modify(1, a, b);
				}
				else if (O[0] == 'S') {
					scanf("%d %d", &a, &b);
					modify(1, a, -b);
				}
				else if (O[0] == 'Q') {
					scanf("%d %d", &a, &b);
					printf("%d\n", find(1, a, b));
				}
				else {
					break;
				}
			}
		}
	}
	return 0;
}

HDU 1166 敌兵布阵（线段树的初步应用2）

原文：http://blog.csdn.net/llx523113241/article/details/42221561