hdu-1166敌兵布阵

这个题目就是考察线段树的基本用法，我自己打了代码，其实就是照模板来的，大概思想已经弄懂了。用c++不能过，说我超时，就改成c的读入读出，这坑爹的过了。我最爱的c++，你肿么了。。。

这是ac的代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int n,m;
int num[100005];

struct H
{
    int l,r,sum;
}trees[300005];

void build_trees(int jd,int l,int r)
{
    trees[jd].l=l;
    trees[jd].r=r;
    if(l==r)
        {trees[jd].sum=num[l];return ;}
    int mid = (l+r)/2;
    build_trees(jd*2,l,mid);
    build_trees(jd*2+1,mid+1,r);
    trees[jd].sum=trees[jd*2].sum+trees[jd*2+1].sum;
}

void update(int jd,int a,int b)
{
    if(trees[jd].l==trees[jd].r)
        trees[jd].sum+=b;
    else {
        int mid = (trees[jd].l+trees[jd].r)/2;
        if(a<=mid) update(jd*2,a,b);
        else update(jd*2+1,a,b);
        trees[jd].sum=trees[jd*2].sum+trees[jd*2+1].sum;
    }
}

int query(int jd , int l,int r)
{
    if(l<=trees[jd].l&&r>=trees[jd].r)
        return trees[jd].sum;
    int ans=0;
    int mid = (trees[jd].l+trees[jd].r)/2;
    if(l<=mid) ans+=query(jd*2,l,r);
    if(r>mid)  ans+=query(jd*2+1,l,r);
    return ans;
}

int main()
{
    int t;
    int i,j,cas;
    int a,b;
    char st[10];
    scanf("%d",&t);
    for(cas=1;cas<=t;cas++)
    {
        memset(num,0,sizeof num);
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        build_trees(1,1,n);
        printf("Case %d:\n",cas);
        for(;;)
        {
            scanf("%s",st);
            if(st[0]==‘E‘)
                break;
            scanf("%d%d",&a,&b);
            if(st[0]==‘A‘)
                update(1,a,b);
            if(st[0]==‘S‘)
                update(1,a,-b);
            if(st[0]==‘Q‘)
                printf("%d\n",query(1,a,b));
        }

    }
    return 0;
}

这是没ac的代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int n,m;
int num[100005] = {0};

struct H
{
    int l,r,sum;
}trees[300005];

void build_trees(int jd,int l,int r)
{
    trees[jd].l=l;
    trees[jd].r=r;
    if(l==r)
        {trees[jd].sum=num[l];return ;}
    int mid = (l+r)/2;
    build_trees(jd*2,l,mid);
    build_trees(jd*2+1,mid+1,r);
    trees[jd].sum=trees[jd*2].sum+trees[jd*2+1].sum;
}

void update(int jd,int a,int b)
{
    if(trees[jd].l==trees[jd].r)
        trees[jd].sum+=b;
    else {
        int mid = (trees[jd].l+trees[jd].r)/2;
        if(a<=mid) update(jd*2,a,b);
        else update(jd*2+1,a,b);
        trees[jd].sum=trees[jd*2].sum+trees[jd*2+1].sum;
    }
}

int query(int jd , int l,int r)
{
    if(l<=trees[jd].l&&r>=trees[jd].r)
        return trees[jd].sum;
    int ans=0;
    int mid = (trees[jd].l+trees[jd].r)/2;
    if(l<=mid) ans+=query(jd*2,l,r);
    if(r>mid)  ans+=query(jd*2+1,l,r);
    return ans;
}

int main()
{
    int l,r;
    cin >> m;
    char s[10];
   for(int j = 1;j <= m;j++)
    {
        cin >> n;
        for(int i = 1;i <= n;i++)
        {

            cin >> num[i];
        }
       build_trees(1,1,n);
    cout << "Case " << j << endl;
        while(1)
        {

            cin >> s;
            if(s[0] == ‘E‘)
            {

                break;
            }
            else if(s[0] == ‘Q‘)
            {
                cin >> l >> r;
                int ans = query(1,l,r);
                cout << ans << endl;
            }
            else if(s[0] == ‘A‘)
            {

                cin >> l >> r;
                update(1,l,r);
            }
            else if(s[0] == ‘S‘)
            {

                cin >> l>> r;
                update(1,l,-r);
            }
        }
    }

    return 0;
}

hdu-1166敌兵布阵