【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;
}