【线段树】hdu1166敌兵布阵

来源：互联网发布：软件部门英文缩写编辑：程序博客网时间：2024/04/28 13:44
/*水水的线段树点修改:----------------------------------------------------------------void build(int l,int r,int o)建树{    int mid = (l + r) / 2;    a[o].left = l;    a[o].right = r;    a[o].num = 0;    if(a[o].left == a[o].right)到达叶子节点        return ;    build(l, mid, lc);向左走    build(mid + 1, r, rc);向右走}--------------------------------------------------------------------------------void update(int l, int r, int o)//L营增加R个人{    if(a[o].left <= l && a[o].right >= l)在该区间则更新num        a[o].num += r;    if(a[o].left == a[o].right)叶子结点返回        return ;    int mid = (a[o].left + a[o].right) / 2;    if(l <= mid)        update(l, r, lc);左走    else        update(l, r, rc);右走}-----------------------------------------------------------------------------------void query(int l, int r, int o)//在id处加L到R的和{    int mid = (a[o].left + a[o].right) / 2;    if(a[o].left == l && a[o].right == r)找到区间更新sum的值    {        sum += a[o].num;        return ;    }    if(r <= mid)        query(l, r, lc);左走    else if(l > mid)        query(l, r, rc);右走    else    {        query(l, mid, lc);        query(mid+1, r, rc);    }}-------------------------------------------------------------------------------------*/#include <iostream>#include <cstdio>#include <cstring>#define INF 0x3f3f3f3f#define lc o*2#define rc o*2+1using namespace std;const int N = 50050*3;struct node{    int left,right;    int num;}a[N];int n,sum;void build(int l,int r,int o){    int mid = (l + r) / 2;    a[o].left = l;    a[o].right = r;    a[o].num = 0;    if(a[o].left == a[o].right)        return ;    build(l, mid, lc);    build(mid + 1, r, rc);}void update(int l, int r, int o)//L营增加R个人{    if(a[o].left <= l && a[o].right >= l)        a[o].num += r;    if(a[o].left == a[o].right)        return ;    int mid = (a[o].left + a[o].right) / 2;    if(l <= mid)        update(l, r, lc);    else        update(l, r, rc);}void query(int l, int r, int o)//在id处加L到R的和{    int mid = (a[o].left + a[o].right) / 2;    if(a[o].left == l && a[o].right == r)    {        sum += a[o].num;        return ;    }    if(r <= mid)        query(l, r, lc);    else if(l > mid)        query(l, r, rc);    else    {        query(l, mid, lc);        query(mid+1, r, rc);    }}int main(){    //freopen("input.txt","r",stdin);    int T;    int kase = 1;    int num;    int l,r;    char s[10];    cin>>T;    while(T--)    {        sum = 0;        printf("Case %d:\n",kase++);        scanf("%d",&n);        build(1, n, 1);        for(int i = 1; i <= n; i++)        {            scanf("%d",&num);            update(i, num, 1);        }        while(scanf("%s",s))        {            if(s[0] == 'E')                break;            if(s[0] == 'Q')            {                scanf("%d%d",&l,&r);                query(l, r, 1);                printf("%d\n",sum);                sum = 0;            }            if(s[0] == 'A')            {                scanf("%d%d",&l,&r);                update(l, r, 1);            }            if(s[0] == 'S')            {                scanf("%d%d",&l,&r);                update(l, -r, 1);            }        }    }    return 0;}
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