【codeforces 500E】New Year Domino
2017-10-04 18:44
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【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 ,l ; vector <pii> v ; map <int,int> dic; vector <int> t; int ans ,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; }
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