hdu-1166敵兵布陣

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這個題目就是考察線段樹的基本用法，我自己打了代碼，其實就是照模板來的，大概思想已經弄懂了。用c++不能過，說我逾時，就改成c的讀入讀出，這坑爹的過了。我最愛的c++，你腫麼了。。。

這是ac的代碼：

#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;}

這是沒ac的代碼：

#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;}