hdu 1166 敌兵布阵(树状数组 | 线段树）

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题目链接：hdu 1166 敌兵布阵

题目大意：略。

解题思路：

方法1：树状数组的水题。这篇题解很好，适合初学树状数组的人。题解连接

#include <stdio.h>#include <string.h>const int N = 50005;const int M = 105;int n, v[N];void add(int x, int val) {while (x <= n) {v[x] += val;x += (x & (-x));}}void init() {memset(v, 0, sizeof(v));int a;scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &a);add(i, a);}}int sum(int x) {int ans = 0;while (x > 0) {ans += v[x];x -= (x & (-x));}return ans;}void solve() {int x, y;char order[M];while (scanf("%s", order) == 1 && strcmp(order, "End") ) {scanf("%d%d", &x, &y);if (order[0] == 'A') add(x, y);else if (order[0] == 'S') add(x, -y);else printf("%d\n", sum(y) - sum(x-1));}}int main() {int cas;scanf("%d", &cas);for (int i = 1; i <= cas; i++) {printf("Case %d:\n", i);init();solve();}return 0;}

方法二：今天看了一下线段树，这题也可以用线段树做，更新点查询区间。

#include <stdio.h>#include <string.h>const int N = 50005;struct node {int l, r, v;}s[N*3];void build(int l, int r, int f) {if (l != r) {int mid = (l + r) / 2;build(l, mid, f*2);build(mid+1, r, f*2+1);}s[f].l = l; s[f].r = r; s[f].v = 0;}void insert(int l, int r, int val, int f) {if (s[f].l != s[f].r) {int mid = (s[f].l + s[f].r) / 2;if (l <= mid && r > mid) { insert(l, mid, val, f*2); insert(mid+1, r, val, f*2+1); }else if (l <= mid && r <= mid) insert(l, r, val, f*2);else insert(l, r, val, f*2+1);}s[f].v += val;}int query(int l, int r, int f) {if (l == s[f].l && r == s[f].r) return s[f].v;int mid = (s[f].l + s[f].r) / 2;if (l <= mid && r > mid) return query(l, mid, f*2) + query(mid+1, r, f*2+1);else if (l <= mid && r <= mid) return query(l, r, f*2);else return query(l, r, f*2+1);}void init () {int n, a;scanf("%d", &n);build(1, n, 1);for (int i = 1; i <= n; i++) {scanf("%d", &a);insert(i, i, a, 1);}}void solve () {int a, b;char o[20];while (scanf("%s", o), strcmp(o, "End")) {scanf("%d%d", &a, &b);if (o[0] == 'A') insert(a, a, b, 1);else if (o[0] == 'S') insert(a, a, -b, 1);else printf("%d\n", query(a, b, 1));}}int main () {int cas;scanf("%d", &cas);for (int i = 1; i <= cas; i++) {printf("Case %d:\n", i);init();solve();}return 0;}

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