标签：acm   hdu   

题意  中文

动态区间和问题   只会更新点  最基础的树状数组 线段树的应用

树状数组代码

#include <bits/stdc++.h>
using namespace std;
const int N = 50005;
int c[N], n, m;

void add(int p, int x)
{
    while(p <= n)
        c[p] += x, p += p & -p;
}

int getSum(int p)
{
    int ret = 0;
    while(p > 0)
        ret += c[p], p -= p & -p;
    return ret;
}

int main()
{
    int u, v, t, cas;
    char op[20];
    scanf("%d", &cas);
    for(int k = 1; k <= cas; ++k)
    {
        printf("Case %d:\n", k);
        scanf("%d", &n);
        memset(c, 0, sizeof(c));
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &t);
            add(i, t);
        }

        while(scanf("%s", op), op[0] != 'E')
        {
            scanf("%d%d", &u, &v);
            if(op[0] == 'Q')
                printf("%d\n", getSum(v) - getSum(u - 1));
            else add(u, op[0] == 'A' ? v : -v);
        }
    }
    return 0;
}

#include <bits/stdc++.h>
#define lc p<<1,s,mid
#define rc p<<1|1,mid+1,e
#define mid ((s+e)>>1)
using namespace std;
const int N = 50005;
int sum[N * 4];

void pushup(int p)
{
    sum[p] = sum[p << 1] + sum[p << 1 | 1];
}

void build(int p, int s, int e)
{
    if(s == e)
    {
        scanf("%d", &sum[p]);
        return;
    }
    build(lc), build(rc);
    pushup(p);
}

void update(int p, int s, int e, int a, int b)
{
    if(s == e && e == a)
    {
        sum[p] += b;
        return;
    }
    if( a <= mid) update(lc, a, b);
    else update(rc, a, b);
    pushup(p);
}

int query(int p, int s, int e, int l, int r)
{
    if(s >= l && e <= r) return sum[p];
    if(r <= mid) return query(lc, l, r);
    if(l > mid) return query(rc, l, r);
    return query(lc, l, mid) + query(rc, mid + 1, r);
}

int main()
{
    int cas, n, a, b;
    char c[20];
    scanf("%d", &cas);
    for(int k = 1; k <= cas; ++k)
    {
        printf("Case %d:\n", k);
        scanf("%d", &n);
        build(1, 1, n);
        while(scanf("%s", c), c[0] != 'E')
        {
            scanf("%d%d", &a, &b);
            if(c[0] == 'Q') printf("%d\n", query(1, 1, n, a, b));
            else if(c[0] == 'A') update(1, 1, n, a, b);
            else update(1, 1, n, a, -b);
        }
    }
    return 0;
}

敌兵布阵

1
10
1 2 3 4 5 6 7 8 9 10
Query 1 3
Add 3 6
Query 2 7
Sub 10 2
Add 6 3
Query 3 10
End 

Case 1:
6
33
59

HDU 1166 敌兵布阵 (树状数组·线段树)

标签：acm   hdu   

原文地址：http://blog.csdn.net/acvay/article/details/45040427