【基础线段树】hdu1166敌兵布阵

题意：给n个兵营的人数，有操作：

Add i j: 给第i个兵营加j人

Sub i j: 第i个兵营减j人

Query i j: 查询第i个到第j个(区间[i,j])的总人数

End: 退出操作.

首先注意到Add和Sub是一样的，Sub i j我们可以用操作Add i -j等价替换掉。于是有两种操作：单点修改，区间查询。因此可以用树状数组简单实现。自然也可以用线段树。

树状数组实现：

#include <cstdio>#include <cstring>const int MAX = 50050;int s[MAX];int lb(int x) {    return x&(-x);}/* œÚµãxÔöŒÓval */void update(int x, int val) {    while (x <= MAX) {        s[x] += val;        x += lb(x);    }}int query(int x) {    int sum = 0;    while (x > 0) {        sum += s[x];        x -= lb(x);    }    return sum;}int main() {    char op[10];    int T, n, a, b;    register int i;    scanf(" %d", &T);    while (T--) {        scanf(" %d", &n);        memset(s, 0, sizeof(s));        for (i = 1; i <= n; ++i) {            scanf(" %d", &a);            update(i, a);        }        static int cas = 0;        printf("Case %d:\n", ++cas);        while (~scanf(" %s", op) && strcmp(op, "End")) {            scanf(" %d %d", &a, &b);            if (strcmp(op, "Query") == 0)                 printf("%d\n", query(b)-query(a-1));            else if (strcmp(op, "Add") == 0)                update(a, b);            else                update(a, -b);        }    }    return 0;}

线段树：

//所有区间表示a~b都为[a,b]，闭区间 #include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <iostream>using namespace std;const int MAX = 50007;int po[MAX];struct node {    int sum; //区间总人数} a[MAX << 2];inline int L(const int& root) {    return root << 1;}inline int R(const int& root) {    return root << 1 | 1;}void build(int root, int left, int right) {    if (left > right) {        return;    } else if (left == right) {        a[root].sum = po[left];    } else {        int mid = (left + right) >> 1;        build(L(root), left, mid);        build(R(root), mid + 1, right);        a[root].sum = a[L(root)].sum + a[R(root)].sum;    }}//给pos加num，num<0时相当于减 void modify(int root, int left, int right, int pos, int num) {    if (pos == left && pos == right) {        a[root].sum += num;    } else if (pos < left || pos > right) {        return;    } else {        int mid = (left + right) >> 1;        modify(L(root), left, mid, pos, num);        modify(R(root), mid + 1, right, pos, num);        a[root].sum = a[L(root)].sum + a[R(root)].sum;    }}int query(int root, int left, int right, int s, int e) {    if (s <= left && e >= right) {        return a[root].sum;    } else if (e < left || s > right) {        return 0;    } else {        int mid = (left + right) >> 1;        return query(L(root), left, mid, s, e) + query(R(root), mid + 1, right, s, e);    }}inline int read() {    char ch;    while ((ch = getchar()) < '0' || ch > '9');    int x = ch - '0';    while ((ch = getchar()) >= '0' && ch <= '9') {        x = (x << 3) + (x << 1) + ch - '0';    }    return x;}int main() {    int T = read();    //scanf(" %d", &T);    while (T--) {        int N = read();        //scanf(" %d", &N);        for (int i = 1; i <= N; ++i) {            po[i] = read();            //scanf(" %d", po + i);        }                build(1, 1, N);                char s[20];        int left, right;        static int __ = 0;        printf("Case %d:\n", ++__);        while (~scanf(" %s", s) && s[0] != 'E') {            left = read();            right = read();            //scanf(" %d %d", &left, &right);            switch(s[0]) {                case 'A':                    modify(1, 1, N, left, right);                    break;                case 'S':                    modify(1, 1, N, left, -right);                    break;                case 'Q':                    printf("%d\n", query(1, 1, N, left, right));                    break;                default:                    break;            }        }    }    return 0;}