hdu 1166 敌兵布阵(线段树、树状数组)

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1166

做的第一题线段树的题（也不全对，之前照模板写了一道st），但是考虑到线段树更新更灵活，而且一般情况下logN不会超时，所以老实打好基本功吧。

WA了一次，因为update的时候，用原数组加减给定的值去更新，但是正确的应该是segTree[rt] += val，索性参考网上的把原始数组也给省略了，有点强大。

当然有树状数组更简单，不过做专题还是好好做，之后补上。

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>#include <queue>using namespace std;const int maxn = 50005;const int oo = 0xfffffff;#define L_CHILD rt<<1, begin, mid#define R_CHILD rt<<1|1, mid + 1, endint segTree[maxn<<2];char s[10];int left, right, idx, val, n, t;void pushUp(int rt){    segTree[rt] = segTree[rt<<1] + segTree[rt<<1|1];}void build(int rt, int begin, int end){    if(begin == end) {        scanf("%d", segTree + rt);        return ;    }    int mid = begin + end >> 1;    build(L_CHILD);    build(R_CHILD);    pushUp(rt);}int query(int rt, int begin, int end, int left, int right){    if(left > end || right < begin) {        return 0;    }    int res = 0;    if(left <= begin && end <= right) {        return segTree[rt];    }    int mid = begin + end >> 1;    res += query(L_CHILD, left, right);    res += query(R_CHILD, left, right);    return res;}void update(int rt, int begin, int end, int idx, int val){    if(begin == end) {        segTree[rt] += val;        return ;    }    int mid = begin + end >> 1;    if(idx <= mid) {        update(L_CHILD, idx, val);    }    else {        update(R_CHILD, idx, val);    }    pushUp(rt);}int main(){    int left, right, idx, val, n;    scanf("%d", &t);    for (int i = 1; i <= t; i ++) {        scanf("%d", &n);        build(1, 1, n);        printf("Case %d:\n", i);        while (scanf("%s", s), s[0] != 'E') {            if(s[0] == 'Q') {                scanf("%d%d", &left, &right);                printf("%d\n", query(1, 1, n, left, right));            }            else if(s[0] == 'A') {                scanf("%d%d", &idx, &val);                update(1, 1, n, idx, val);            }            else if(s[0] == 'S') {                scanf("%d%d", &idx, &val);                update(1, 1, n, idx, -val);            }        }    }    return 0;}

今天看了下树状数组，发现在求和问题上比线段树方便多了，于是再把这题写了一下，相比线段树，这个理解起来完全就不是问题了。

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>#include <queue>using namespace std;const int maxn = 50005;const int oo = 0xfffffff;int bitTree[maxn + 1];char s[10];int n, t;int lowbit(int i) {    return i & (-i);}int add(int i, int val) {    while (i <= n) {        bitTree[i] += val;        i += lowbit(i);    }}int sum(int i) {    int s = 0;    while (i > 0) {        s += bitTree[i];        i -= lowbit(i);    }    return s;}int main(){    int left, right, idx, val;    scanf("%d", &t);    for (int i = 1; i <= t; i ++) {        memset(bitTree, 0 ,sizeof(bitTree));        scanf("%d", &n);        for (int j = 1; j <= n; j ++) {            scanf("%d", &val);            add(j, val);        }        printf("Case %d:\n", i);        while (scanf("%s", s), s[0] != 'E') {            if(s[0] == 'Q') {                scanf("%d%d", &left, &right);                printf("%d\n", sum(right) - sum(left-1));            }            else if(s[0] == 'A') {                scanf("%d%d", &idx, &val);                add(idx, val);            }            else if(s[0] == 'S') {                scanf("%d%d", &idx, &val);                add(idx, -val);            }        }    }    return 0;}

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