线段树(单点更新)/树状数组 HDOJ 1166 敌兵布阵

 

题目传送门

 1 /*  2  线段树基本功能：区间值的和，修改某个值  3 */  4 #include <cstdio>  5 #include <cstring>  6 #define lson l, m, rt << 1  7 #define rson m+1, r, rt << 1|1  8  9 const int MAX_N = 50000 + 10; 10 int sum[MAX_N<<2]; 11 12 void pushup(int rt) //杭电大牛：notOnlySuccess 版本 13 { 14 sum[rt] = sum[rt<<1] + sum[rt<<1|1]; 15 } 16 17 void build(int l, int r, int rt) 18 { 19 if (l == r) 20  { 21 scanf ("%d", &sum[rt]); 22 return ; 23  } 24 int m = (l + r) >> 1; 25  build (lson); 26  build (rson); 27  pushup (rt); 28 } 29 30 void update(int p, int add, int l, int r, int rt) 31 { 32 if (l == r) 33  { 34 sum[rt] += add; 35 return ; 36  } 37 int m = (l + r) >> 1; 38 if (p <= m) update (p, add, lson); 39 else update (p, add, rson); 40  pushup (rt); 41 } 42 43 int query(int ql, int qr, int l, int r, int rt) 44 { 45 if (ql <= l && r <= qr) 46  { 47 return sum[rt]; 48  } 49 int m = (l + r) >> 1; 50 int ans = 0; 51 if (ql <= m) ans += query (ql, qr, lson); 52 if (qr > m) ans += query (ql, qr, rson); 53 54 return ans; 55 } 56 57 int main(void) //HDOJ 1166 敌兵布阵 58 { 59 //freopen ("inA.txt", "r", stdin); 60 int t, n, cas, ans; 61 int b, c; 62 char s[10]; 63 64 while (~scanf ("%d", &t)) 65  { 66 cas = 0; 67 while (t--) 68  { 69 scanf ("%d", &n); 70 build (1, n, 1); 71 printf ("Case %d:\n", ++cas); 72 while (~scanf ("%s", &s)) 73  { 74 if (strcmp (s, "End") == 0) break; 75 scanf ("%d%d", &b, &c); 76 if (strcmp (s, "Query") == 0) 77  { 78 ans = query (b, c, 1, n, 1); 79 printf ("%d\n", ans); 80  } 81 else if (strcmp (s, "Add") == 0) 82  { 83 update (b, c, 1, n, 1); 84  } 85 else if (strcmp (s, "Sub") == 0) 86  { 87 update (b, -c, 1, n, 1); 88  } 89  } 90  } 91  } 92 93 return 0; 94 }

 1 /*  2  树状数组  3 */  4 #include <cstdio>  5 #include <cstring>  6  7 const int MAX_N = 50000 + 10;  8 int a[MAX_N];  9 int n, num; 10 11 int lowbit(int t) //求 2^k 12 { 13 //return t & (t ^ (t - 1)); 14 return t & (-t); 15 } 16 17 void plus(int num, int x) //第i个增加num个人 18 { 19 while (x <= n) 20  { 21 a[x] += num; 22 x += lowbit (x); 23  } 24 } 25 26 int sum(int x) //求前x项的和 27 { 28 int sum = 0; 29 while (x > 0) 30  { 31 sum += a[x]; 32 x -= lowbit(x); 33  } 34 return sum; 35 } 36 37 int main(void) 38 { 39 //freopen ("inA.txt", "r", stdin); 40 41 int t; 42 char op[100]; 43 44 scanf ("%d", &t); 45 int k = t; 46 47 while (k--) 48  { 49 memset (a, 0, sizeof (a)); 50 scanf ("%d", &n); 51 for (int i=1; i<=n; ++i) 52  { 53 scanf ("%d", &num); 54  plus (num, i); 55  } 56 int r = 1; 57 while (scanf ("%s", &op), op[0] != 'E') 58  { 59 int a, b; 60 scanf ("%d%d", &a, &b); 61 switch (op[0]) 62  { 63 case 'A': 64  plus (b, a); 65 break; 66 case 'S': 67 plus (-b, a); 68 break; 69 case 'Q': 70 if (r) 71 printf ("Case %d:\n", t - k); 72 r = 0; 73 printf ("%d\n", sum (b) - sum (a-1)); 74 break; 75  } 76  } 77  } 78 79 return 0; 80 }

树状数组