HDOJ-1166 敌兵布阵

来源：互联网发布：御彩轩时时彩计划软件编辑：程序博客网时间：2024/06/06 09:50

这道题数据量很大，用朴素的方法做每个操作的时间复杂度为o(n)会超时，如果用线段树则每个操作为o(logn).

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <climits>#include <set>using namespace std;#define maxn 50005struct node{    int l, r, s;}T[maxn*3];int p[maxn];int Build(int n, int l, int r){    T[n].l = l;    T[n].r = r;    if(l == r){        return T[n].s = p[l];    }    int mid = (l + r) / 2;    return T[n].s = Build(n<<1, l, mid) + Build(n<<1|1, mid+1, r);}void Update(int n, int m, int k){    if(T[n].l == T[n].r){       T[n].s = k;       return ;    }    int mid = (T[n].l + T[n].r) / 2;    if(m <= mid)        Update(n<<1, m, k);    else        Update(n<<1|1, m, k);    T[n].s = T[n<<1].s + T[n<<1|1].s;}void Query(int n, int l, int r, int &sum){    if(T[n].l == l && T[n].r == r){        sum += T[n].s;        return ;    }    int mid = (T[n].l + T[n].r) / 2;    if(r <= mid)        Query(n<<1, l, r, sum);    else if(l > mid)        Query(n<<1|1, l, r, sum);    else {        Query(n<<1, l, mid, sum);        Query(n<<1|1, mid+1, r, sum);    }}int main(){  //  freopen("in.txt", "r", stdin);    int t, cas = 0;    cin >> t;    while(t--)    {        printf("Case %d:\n", ++cas);        string s;        int n, a, b;        cin >> n;        for(int i = 1; i <= n; i++)            scanf("%d", p+i);        Build(1, 1, n);        while(cin >> s && s[0] != 'E')        {            if(s[0] == 'Q')            {                int sum = 0;                cin >> a >> b;                Query(1, a, b, sum);                cout << sum << endl;            }            else if(s[0] == 'A')            {                cin >> a >> b;                p[a] += b;                Update(1, a, p[a]);            }            else if(s[0] == 'S')            {                cin >> a >> b;                p[a] -= b;                Update(1, a, p[a]);            }        }    }    return 0;}

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