线段树 基础单点更新 敌兵布阵

题：敌兵布阵

标准线段树模板代码：

#include<cstdio>#include<cstring>const int maxn = 500000 + 10;struct Node{    int left, right, count;}node[maxn];int a[maxn];/**************************************************建树****************************i是区间序号**************l是区间i左边界，r是区间i右边界***从1到n开始建树，直到区间长度为1***即l=r时，结束。count记录区间和**************************************/void maketree(int l, int r, int i){    node[i].left = l;    node[i].right = r;    if(l == r){        node[i].count = a[l];        return ;    }    int m = (l + r)/2;    maketree(l, m, 2*i);    maketree(m + 1, r, 2*i + 1);    node[i].count = node[2*i].count + node[2*i + 1].count;}/*********************************************************更新********************i区间序号，x要更新的点。y要更新的值**********flag判断更新方式*****************************************************/void updatetree(int i, int x, int y, int flag){    int l = node[i].left;    int r = node[i].right;    int m = (l + r)/2;    if(r == l){        if(flag)            node[i].count += y;        else            node[i].count -= y;        return;    }    if(x <= m)        updatetree(2*i, x, y, flag);    else        updatetree(2*i + 1, x, y, flag);    if(flag)        node[i].count += y;    else        node[i].count -= y;    return;}/**************************************************查询****************************************************/int querytree(int l, int r, int i){    int m = (node[i].left + node[i].right)/2;    if(node[i].right <= r && node[i].left >= l) return node[i].count;    int  ans = 0;    if(r <= m)        return querytree(l, r, 2*i);    else        if(l > m)            return querytree(l, r, 2*i + 1);        else            return querytree(l, m, 2*i) + querytree(m + 1, r, 2*i + 1);}int main(){    int T, n;    char str[20];    scanf("%d", &T);    for(int i = 1; i <= T; i++){        printf("Case %d:\n", i);        scanf("%d", &n);        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);        maketree(1, n, 1);        int x, y;        while(scanf("%s", str)){            if(str[0] == 'E') break;            scanf("%d%d", &x, &y);            if(str[0] == 'Q') printf("%d\n", querytree(x, y, 1));            else                if(str[0] == 'A') updatetree(1, x, y, true);                    else updatetree(1, x, y, false);        }    }    return 0;}

优美的线段树代码：

#include <cstdio>/*****************************灵活的使用宏定义******************************/#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn = 55555;int sum[maxn<<2];void PushUP(int rt) {sum[rt] = sum[rt<<1] + sum[rt<<1|1];}/***************************************************建树*************** 此处并没有使用结构体，只是记录了区间和sum，但l，r与区间序号紧密关联***********************************/void build(int l,int r,int rt) {if (l == r) {scanf("%d",&sum[rt]);return ;}int m = (l + r) >> 1;build(lson);build(rson);PushUP(rt);}/*************************************** *****************更新***************** 此处没用标记更新方式（加或减），巧妙 地在调用时处理了正负号，减少函数参数***************************************/void update(int p,int add,int l,int r,int rt) {if (l == r) {sum[rt] += add;return ;}int m = (l + r) >> 1;if (p <= m) update(p , add , lson);else update(p , add , rson);PushUP(rt);}/*********************************************************查询****************** 巧妙地引用了变量ret，减少了对m的讨论***************************************/int query(int L,int R,int l,int r,int rt) {if (L <= l && r <= R) {return sum[rt];}int m = (l + r) >> 1;int ret = 0;if (L <= m) ret += query(L , R , lson);if (R > m) ret += query(L , R , rson);return ret;}int main() {int T , n;scanf("%d",&T);for (int cas = 1 ; cas <= T ; cas ++) {printf("Case %d:\n",cas);scanf("%d",&n);build(1 , n , 1);char op[10];while (scanf("%s",op)) {if (op[0] == 'E') break;int a , b;scanf("%d%d",&a,&b);if (op[0] == 'Q') printf("%d\n",query(a , b , 1 , n , 1));else if (op[0] == 'S') update(a , -b , 1 , n , 1);else update(a , b , 1 , n , 1);}}return 0;}

上述两种代码思路相同，只是代码风格不同，运行时间，占用内存还是相同的。此题只涉及单点更新和区间求和，所以可以用树状数组求解，代码更简洁，运行速度更快。但树状数组可以求区间和，无法求出区间最值，通用解法仍是用线段树求解。
树状数组的代码：

#include<cstdio>#include<cstring>using namespace std;const int maxn = 50000 + 10;int len, a[maxn];char str[50];int lowbit(int x){    return x&(-x);}/******************************更新********************************/void update(int i, int v){    while(i <= len){        a[i] += v;        i += lowbit(i);    }}/*****************************求和******************************/int sum(int i){    int sum = 0;    while(i > 0){        sum += a[i];        i -= lowbit(i);    }    return sum;}int main(){    int T, v;    scanf("%d", &T);    for(int i = 1; i <= T; i++){        memset(a, 0, sizeof(a));        scanf("%d", &len);        for(int j = 1; j <= len; j++){            scanf("%d", &v);            update(j, v);        }        printf("Case %d:\n", i);        while(scanf("%s", str)){            if(str[0] == 'E') break;            int x, y;            scanf("%d%d", &x, &y);            if(str[0] == 'A') update(x, y);            else                if(str[0] == 'S') update(x, -y);                    else printf("%d\n", sum(y)-sum(x-1));        }    }    return 0;}