Hello to all. I've (Trofimenko:)) solved this problem. In my program I used constant MAXM=5000. I've got WA5. When I changed MAXM to 6000 I've got WA too. But when I changed maxm to 60000 I've got AC. It's so stupid... The set contains at least one and at most 99999 segments! What's means of your MAXM??? The number of segments? If yes,you are wrong without doubt! No, I was using it to limit M which should be <=5000 "covers segment [0, M] completely" means: 0 1 2 3 does not cover [0,3] completely 0 1 1 3 covers [0,3] completely Thank you for your tests! Tested. Still WA#2 Maybe you printed “No solution[space]” instead of “No solution” like me :)) This is so dumb. But thanks for clarification. I don't understand why they did not mention it in problem statement. Or I might have not noticed it. Could anybody help me? I have WA at 3th test .I tried many test and answers were correct .I also tried this test and answer was correct: 5 -5 0 0 1 1 2 2 3 4 5 3 4 10 49000 0 0 Hey, this is the test 16 - it was a long path to get there. What can possibly go wrong? I don't understand why my prog don't work Can you give me some data or advice Edited by author 20.03.2007 14:04 Edited by author 20.03.2007 14:04 10 0 9 0 0 I had "Output limit exceeded" on this test, but it actually was WA12. It works. Thanks man, appreciate it! <3 i have tried all cases and it seems correct but cant pass test 13. Can someone help me what is test 13??? Английский я знаю плохо, поэтому sorry:) У меня тоже был WA 12. Попробуй тест: 1 49999 50000 0 0 TY so much <3 :(( Английский я знаю плохо, поэтому sorry:) У меня тоже был WA 12. Попробуй тест: 1 49999 50000 0 0 Edited by author 25.03.2022 08:26I have solved this problem by using an easy greedy algorithm. MAIN: you need to use an idea of 2 pointers(of course you need to sort array) and take the rightest segment. Left part of segment need to be <= x(another pointer) The solution below: void solve() { int m; cin >> m; vector< pair<int, int> > pi; int a, b; cin >> a >> b; while (a || b) { pi.pb({a, b}); cin >> a >> b; } sort(all(pi)); int cnt = 0, j = 0; vector< pair<int, int> > ans; j = 0; int x = 0; for (; j < sz(pi) && x < m; j++) { int k = j; int mx = x; pair<int, int> par = {-INF, -INF}; while (k < sz(pi) && pi[k].ff <= x) { if (mx < pi[k].ss) { mx = pi[k].ss; par = pi[k]; } k++; } if (par.ff == -INF) { cout << "No solution\n"; return; } x = mx; ans.pb(par); if (k > j) j = k - 1; } if (x < m) { cout << "No solution\n"; return; } cout << sz(ans) << '\n'; for (auto &i : ans) { cout << i.ff << ' ' << i.ss << '\n'; } } always choose segments with rightest right points dp[i] - минимальное количество отрезков, если последний отрезок заканчивался в позиции i. ответ dp[m]. What can you tell me about this test? I'm having WA all the time Same question. I tried all tests of this problem's forum and still got WA3. Edited by author 25.02.2013 21:48 just try 5 -5 0 0 1 1 2 2 3 4 5 3 4 10 49000 0 0 There is full solution!!! Try thinking yourself or using hints before watching it and use visual c++ 2017 . . . . . . . . . . . . . . . . . . . //#include <bits/stdc++.h> #include <iostream> #include <vector> #include <algorithm> #define int long long #define pb push_back using namespace std; const int N = 50000; pair<int, int> dp[N], right_end[N]; vector<pair<int, int> > v; vector<int> ans; unsigned main() { for(int i = 0; i < N; i++) { right_end[i].first = -50001; right_end[i].second = -50001; } int m; cin >> m; int l, r; int k = 0; while(true) { cin >> l >> r; if(l > m || r < 0) continue; if(l == 0 && r == 0) break;
v.pb({l, r});
if(r > right_end[l].first) { right_end[l].first = r; right_end[l].second = k; } k++; }
int i = 0; for(auto it: v) { if(it.first <= 0 && it.second > 0) { if(it.second > dp[0].first) { dp[0].first = it.second; dp[0].second = i; } } i++; } for(int i = 1; i <= m; i++) { if(dp[i - 1].first >= right_end[i].first) { dp[i].first = dp[i - 1].first; dp[i].second = dp[i - 1].second; } else { dp[i].first = right_end[i].first; dp[i].second = right_end[i].second; } } for(int i = 0; i <= m; i++) { if(dp[i].first < i) { cout << "No solution" << endl; return 0; } } int pos = 0; while(pos < m) { ans.pb(dp[pos].second); if(dp[pos].first == pos) { cout << "No solution" << endl; return 0; } pos = dp[pos].first; } cout << ans.size() << endl; for(auto it: ans) { cout << v[(int)it].first << " " << v[(int)it].second << endl; } } Edited by author 09.02.2020 01:50 Could anybody help me? I have WA at 4th test! u may try this test: 2 1 2 0 1 0 0 so output should be: 2 0 1 1 2 when WA is 2 1 2 0 1 This test helped me: 4 0 4 -5 0 3 4 -4 4 0 0 Answer is: 1 -4 4 Is the answer : 1 0 4 right? i have wa 4 too,and i am confused about the output principles. This test helped me: 10 -5 1 1 14 -5 2 2 3 3 19 0 0 The answer is: 2 -5 1 1 14 Now, let's fight with WA5 :D why not this: 2 -5 2 1 14 ? Actually, what if there are several possibilites to cover? I think that it does not matter which one you choose, as long as it has a minimum number of intervals. I guess the judge just tests whether the intervals you give cover the large interval. Best regards, Erik Edited by author 25.02.2013 04:56 This test helped me too. At last got AC. My solution is similar to task 1987. Edited by author 05.09.2019 16:47 1) greedy works here, although it's not that much obvious how *correct* greedy should look like 2) 5th test case, that didn't pass for me: 12 -3 10 -2 8 -1 16 0 0 Edited by author 17.02.2018 04:30 The problem is quite evil. The problem is actually very easy. It is hard to understand that the problem is that easy. Possible solution. First , eliminate all such segments I=[l;r], that l>M or r<0, because of such I's can't appear in the answer. Second, build your dp[]. Your dp[x] contains rightmost right end of all such segments [l;r], that l<=x. Finally, just start from segment whose right end is dp[0], jump to a segment whose right end is dp[dp[0]] and so on. How to build dp[x]? I have used an additional array right_end[y]. Where right_end[y] is the right end of a segment with maximal length from all segments whose left end is y. dp[x] =max(dp[x-1], right_end[x]). Hi! I use greedy algorithms and can't get past first test. Could someone please provide me with some tests? I tried these tests with following results: 1 -1 0 -5 -3 2 5 0 0 ----------- No solution 1 -1 0 0 1 0 0 ----------- 0 1 10 -5 1 1 14 -5 2 2 3 3 19 0 0 --------- -5 2 1 14 5 -5 0 0 1 1 2 2 3 4 5 3 4 10 49000 0 0 --------- 0 1 1 2 2 3 3 4 4 5 4 0 4 -5 0 3 4 -4 4 0 0 ------- -4 4 6 0 5 7 8 0 0 ------- No solution 10 0 9 0 0 ------- No solution you don't print segment's count first line #include <bits/stdc++.h> #define f first #define s second #define pb push_back #define ppb pop_back #define mp make_pair #define pii pair <int, int> #define pll pair <ll, ll> #define ld long double #define ll long long #define ull unsigned ll #define bit(x) __builtin_popcountll(x) #define all(x) x.begin(), x.end() #define sqr(x) ((x) * 1ll * (x)) #define sz(x) (int)x.size() #define purple ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0); #define rep(_i, _from, _to) for (int _i = _from; _i <= _to; _i++) #define per(_i, _from, _to) for (int _i = _from; _i >= _to; _i--) #define nl '\n' #define ioi exit(0); #define _225day "" using namespace std; const int N = 1e6 + 7, mod = 1e9 + 9, inf = 1e9 + 7; const ll linf = (ll)1e18 + 7; const ld eps = 1e-15, pi = 3.141592; const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1}, dy[] = {0, 1, 0, -1, 1, -1, 1, -1}; int n, m, base = 50000; int s[N]; pii a[N]; struct Tree{ int ans, add; } t[N << 2]; inline bool cmp(pii x, pii y){ if (x.f != y.f) return x.f < y.f; } inline bool Check(int pref){ memset(s, 0, sizeof(s)); rep(i, 1, pref) s[a[i].f]++, s[a[i].s + 1]--; rep(i, 0, m + base){ s[i] += s[i - 1]; if (i > base && !s[i]) return 0; } return 1; } inline void Push(int v, int tl, int tr){ if (t[v].add && tl != tr){ t[v + v].add += t[v].add, t[v + v + 1].add += t[v].add; t[v + v].ans += t[v].add, t[v + v + 1].ans += t[v].add; t[v].add = 0; } } inline void Update(int l, int r, int add, int v = 1, int tl = 0, int tr = N){ Push(v, tl, tr); if (l <= tl && tr <= r){ t[v].ans += add; t[v].add += add; return; } if (tl > r || tr < l) return; int tm = tl + tr >> 1; Update(l, r, add, v + v, tl, tm); Update(l, r, add, v + v + 1, tm + 1, tr); t[v].ans = min(t[v + v].ans, t[v + v + 1].ans); } inline int Get(int l, int r, int v = 1, int tl = 0, int tr = N){ Push(v, tl, tr); if (l <= tl && tr <= r) return t[v].ans; if (tl > r || tr < l) return inf; int tm = tl + tr >> 1; return min(Get(l, r, v + v, tl, tm), Get(l, r, v + v + 1, tm + 1, tr)); } int main(){ #ifndef _225day freopen (_225day".in", "r", stdin); freopen (_225day".out", "w", stdout); #endif cin >> m; while (1){ int x, y; cin >> x >> y; if (x == 0 && y == 0) break; a[++n] = {x + base + 1, y + base}; } sort (a + 1, a + 1 + n, cmp); int l = 1, r = n, ans = -1, mid; while (l <= r){ mid = l + r >> 1; if (Check(mid)) ans = mid, r = mid - 1; else l = mid + 1; } if (ans == -1) cout << "No solution", ioi ans = n; rep(i, 1, ans) Update(a[i].f, a[i].s, 1); vector <pii> res; rep(i, 1, ans){ Update(a[i].f, a[i].s, -1); if (!Get(base + 1, m + base)) Update(a[i].f, a[i].s, 1), res.pb({a[i].f, a[i].s}); } sort (all(res), cmp); cout << sz(res) << nl; for (auto i : res) cout << i.f - base - 1 << ' ' << i.s - base << nl; ioi } Same problem :( WA5, WA5, WA5... What causes that? When my sort compare function is: bool cmp(pair<int, int> p1, pair<int, int> p2) { if(p1.first != p2.first) return p1.first < p2.first; else return p1.second >= p2.second; } I get WA#14. And then delete the "=" like : bool cmp(pair<int, int> p1, pair<int, int> p2) { if(p1.first != p2.first) return p1.first < p2.first; else return p1.second > p2.second; } I get AC; Try this test: 5 0 3 3 5 5 7 0 0 AC answer is: 2 0 3 3 5 not: 3 0 3 3 5 5 7 Исправьте слово "соледует" в последнем предложении Результата: "соледует вывести единственную фразу «No solution»". #include <cstdio> #include <climits> #include <cstdlib> #include <algorithm> #include <list> #include <vector> using namespace std; #define PB push_back #define BEG begin() #define MP make_pair #define F first #define S second bool compare(const pair<int,int> &left, const pair<int,int> &right){ return left.F <= right.F; } int main(){ int M; int readInt1, readInt2; vector<pair<int,int> > segments; vector<int> solutionInd; scanf("%d", &M); while(scanf("%d %d", &readInt1, &readInt2) && !(readInt1 == 0 && readInt2 == 0)){ segments.PB(MP(readInt1, readInt2)); } sort(segments.begin(), segments.end(), compare); if(segments.empty() || segments[0].F > 0){ printf("No solution\n"); return 0; }
int currEnd = 0; int segInd = 0; int ansInd = -1; int maxEnd = 0;
while(segInd < segments.size()){ while(segInd < segments.size() && segments[segInd].F <= currEnd){ if(segments[segInd].S > maxEnd){ maxEnd = segments[segInd].S; ansInd = segInd; }
segInd++; } solutionInd.PB(ansInd); currEnd = maxEnd;
if(currEnd >= M || segments[segInd].F > currEnd) break; } if(solutionInd.empty() || currEnd < M){ printf("No solution\n"); } else{ printf("%d \n", solutionInd.size()); for(int i = 0; i<solutionInd.size(); i++){ printf("%d %d\n", segments[solutionInd[i]].F, segments[solutionInd[i]].S); } }
}
Keep getting runtime error(Access violation) even though I can't find anywhere it is possible to go outside the vectors memory space.
Update: The problem was accepted when using visual studio compiler instead of g++. Same thing happened to me. But why is that? Hi, I get WA6 with g++ but get runtime error access violation on test case 14 with Visual Studio 2010 C++. Code : /* * File: main.cpp * Author: Parnami * Created on April 14, 2014, 10:18 PM * Description : TIMUS Online Judge Problem ID : 1303 (DP) */ #include <stdio.h> #include <iostream> #include <algorithm> #include <vector> #include <map> #include <string> #include <queue> #include <cmath> #include <utility> #include <limits.h> using namespace std; inline int fastRead() { int input; char c=0; while (c<33) c=getchar(); input=0; while (c>33) { input=input*10+c-'0'; c=getchar(); } return input; } vector < pair <int,int> > myVec,answer; bool inRange(pair <int,int> cur, int starter) { if(starter>=cur.first && starter<cur.second) return true; return false; } int main(int argc, char** argv) { int m,starter,ender,iter,i,flag; pair <int,int> maxxer,current; m = fastRead(); while(1) { cin>>starter>>ender; if(starter==0&&ender==0) break; if(ender<=0||(starter>=m&&ender>m)) { //Do Nothing } else { //cout<<"Pushing "<<starter<<"-"<<ender<<endl; myVec.push_back(make_pair(starter,ender)); } } if(!myVec.empty()) { sort(myVec.begin(),myVec.end()); iter = 0; starter=0; while(starter<m) { //cout<<starter<<endl; current = myVec[iter]; //cout<<"Yahaan Phuncha"<<endl; flag = 0; while(!inRange(current,starter) && iter!=myVec.size()) { //cout<<"Not in range : "<<current.first<<"-"<<current.second<<endl; iter++; current = myVec[iter]; } if(iter==myVec.size()) { //cout<<"End of vector nowhere to search"<<endl; flag = 1; break; } maxxer = current; iter++; if(iter==myVec.size()) {
} else { while(1) { if(myVec[iter].first>starter) { break; } else { if(myVec[iter].second>maxxer.second) { maxxer = myVec[iter]; } iter++; } } } //cout<<"Pushing into answer "<<maxxer.first<<"-"<<maxxer.second<<endl; answer.push_back(maxxer); starter = maxxer.second; } } else flag=1; if(flag) { cout<<"No solution"<<endl; } else { cout<<answer.size()<<endl; for(i=0;i<answer.size();i++) { cout<<answer[i].first<<" "<<answer[i].second<<endl; } } return 0; } Edited by author 15.04.2014 04:19 |
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