【题目链接】:http://codeforces.com/problemset/problem/500/E

【题意】



有n个多米诺骨牌;

你知道它们的长度;

然后问你,如果把第i骨牌往后推倒,然后要求第i到第j个骨牌(j>i)都倒掉;

问你需要把i..j这里面骨牌总共增高多少单位的长度(输出最小值);

【题解】



从最后一个骨牌开始往前处理;

对于每一个骨牌,把p[i]..p[i]+l[i]全都覆盖;

然后对于询问x[i],y[i];

即查询p[x[i]]..p[y[i]]这个区间里面有多少个空格;

这两个操作都能用线段树完成;

写线段树的时候要写坐标压缩.



【Number Of WA】



6



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0),cin.tie(0)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 2e5+100;

const int MAX_SIZE = 4e5+100;

int n,q,ma;

int p[N],l[N];

vector <pii> v[N];

map <int,int> dic;

vector <int> t;

int ans[N],lazy_tag[MAX_SIZE<<2],sum[MAX_SIZE<<2];

void push_down(int rt,int l,int r)

{

    if (lazy_tag[rt]==0) return;

    lazy_tag[rt] = 0;

    sum[rt] = t[r+1]-t[l];

    if (l!=r)

        lazy_tag[rt<<1] = lazy_tag[rt<<1|1] = 1;

}

void up_data(int L,int R,int l ,int r,int rt)

{

    push_down(rt,l,r);

    if (L <= l && r <= R)

    {

        lazy_tag[rt] = 1;

        push_down(rt,l,r);

        return;

    }

    int m = (l+r)>>1;

    if (L <= m)

        up_data(L,R,lson);

    if (m < R)

        up_data(L,R,rson);

    push_down(rt<<1,l,m);push_down(rt<<1|1,m+1,r);

    sum[rt] = sum[rt<<1] + sum[rt<<1|1];

}

int query(int L,int R,int l,int r,int rt)

{

    //cout <<L<<' '<<R<<' '<<l<<' '<<r<<' '<<endl;

    push_down(rt,l,r);

    if (L<=l && r <= R)

        return sum[rt];

    int m = (l+r)>>1;

    int temp1 = 0,temp2 = 0;

    if (L<=m)

        temp1=query(L,R,lson);

    if (m<R)

        temp2=query(L,R,rson);

    //cout <<temp1+temp2<<endl;

    return temp1+temp2;

}

int main()

{

    ms(sum,0);

    ms(lazy_tag,0);

   // Open();

    Close();//scanf,puts,printf not use

    //init??????

    cin >> n;

    t.pb(-1);

    rep1(i,1,n)

    {

        cin >> p[i] >> l[i];

        if (!dic[p[i]])

        {

            dic[p[i]] = 1;

            t.pb(p[i]);

        }

        if (!dic[p[i]+l[i]])

        {

            dic[p[i]+l[i]] = 1;

            t.pb(p[i]+l[i]);

        }

    }

    sort(t.begin(),t.end());

    ma = int(t.size())-1;

    dic.clear();

    rep1(i,1,ma)

        dic[t[i]] = i;

    cin >> q;

    rep1(i,1,q)

    {

        int x,y;

        cin >>x >> y;

        v[x].pb(mp(y,i));

    }

    rep2(i,n,1)

    {

        int xb1 = dic[p[i]],xb2 = dic[p[i]+l[i]];

        //cout <<xb1<<' '<<xb2<<endl;

        //cout <<xb1<<' '<<xb2<<endl;

        up_data(xb1,xb2-1,1,ma-1,1);

        //if (i==3)

        //{

        //    cout <<xb1<<' '<<xb2<<endl;

        //    return 0;

       // }

        //cout <<dic[p[i]]<<' '<<dic[p[i+1]]<<endl;

        //cout <<xb1<<' '<<xb2<<endl;

        //cout << query(dic[p[i]],dic[p[i+1]]-1,1,ma,1) << endl;

        //return 0;

        int len = v[i].size();

        rep1(j,0,len-1)

        {

            int w = v[i][j].fi,id = v[i][j].se;

            int xb3 = dic[p[w]];

            //cout <<xb1<<' '<<xb3-1<<endl;

            ans[id] = p[w]-p[i]-query(xb1,xb3-1,1,ma-1,1);

            //cout <<p[w]-p[i]<<endl;

            //cout <<query(xb1,xb3-1,1,ma-1,1)<<endl;

        }

    }

    rep1(i,1,q)

        cout << ans[i] << endl;

    return 0;

}