이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define REPO(i,n) for(int (i)=1; (i)<=(int)(n); (i)++)
#define SZ(v) ((int)(v).size())
#define ALL(v) (v).begin(),(v).end()
#define one first
#define two second
typedef long long ll;
typedef pair<int, int> pi;
const int INF = 0x3f2f1f0f;
const ll LINF = 1ll * INF * INF;
struct PT {
int x, y;
PT() {}
PT(int xx, int yy) : x(xx), y(yy) {}
bool operator<(const PT &o) const{return x+y < o.x+o.y;}
};
struct SOL {
priority_queue<int> m; //minus slope
priority_queue<int, vector<int>, greater<int> > p; //plus slope;
ll ans;
SOL() : ans(0) {}
void add(int l, int r) {
m.push(l); p.push(r);
// printf("%d %d add\n", l, r);
while(true) {
int mm = m.top();
int pp = p.top();
if(mm <= pp) return;
p.pop(); m.pop();
ans += abs(pp - mm);
m.push(pp); p.push(mm);
// printf("%d %d pop\n", mm, pp);
// printf("%d %d push\n", pp, mm);
}
}
};
const int MAX_N = 1e5 + 100;
int K, N;
ll Ans = 0;
vector<PT> Ps;
ll DyL[MAX_N], DyR[MAX_N];
int main() {
cin >> K >> N;
REP(i, N) {
char sa[3], sb[3]; int x, y;
scanf("%s%d%s%d", sa, &x, sb, &y);
if(sa[0] != sb[0]) {
Ans += 1;
Ps.push_back(PT(min(x, y), max(x, y)));
}
Ans += abs(x-y);
}
sort(ALL(Ps));
if(K == 1) {
SOL sol;
for(PT &p : Ps) sol.add(p.x, p.y);
Ans += sol.ans * 2;
printf("%lld\n", Ans);
} else {
SOL lsol, rsol;
for(int i=0; i<SZ(Ps); i++) {
PT &p = Ps[i];
lsol.add(p.x, p.y);
DyL[i+1] = lsol.ans * 2;
}
for(int i=SZ(Ps)-1; i>=0; i--) {
PT &p = Ps[i];
rsol.add(p.x, p.y);
DyR[i+1] = rsol.ans * 2;
}
ll nowAns = LINF;
for(int i=0; i<=SZ(Ps); i++)
nowAns = min(nowAns, DyL[i] + DyR[i+1]);
Ans += nowAns;
printf("%lld\n", Ans);
}
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
bridge.cpp: In function 'int main()':
bridge.cpp:53:36: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
scanf("%s%d%s%d", sa, &x, sb, &y);
^
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