【codeforces 500E】New Year Domino

【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;
}