Codeforces 433 Div.2（A、B、C、D）
A. Fraction
暴力遍历1-1000，取组成的真分数比值最大且分子分母gcd为1时更新答案
代码：
#include <stdio.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
int main(void)
{
int n;
while (~scanf("%d", &n))
{
double mm = 0;
int aa, bb;
for (int a = 1; a <= n; ++a)
{
for (int b = a + 1; b <= n; ++b)
{
if (a + b == n && a < b && (a * 1.0 / b) > mm && __gcd(a, b) == 1)
{
aa = a; bb = b;
mm = a * 1.0 / b;
}
}
}
printf("%d %d\n", aa, bb);
}
return 0;
}
B. Maxim Buys an Apartment
最好的情况是尽量放中间，每一个房子都可以造成两个good位置，这样一共有n/3个位置，如果k<=n/3则答案就是k*3，否则把剩余的位置先放到对答案没影响的，再放到对答案影响为1的地方。
代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
int main(void)
{
LL n, k;
while (cin >> n >> k)
{
if (k == 0 || k == n)
puts("0 0");
else
{
if (n == 2)
puts("1 1");
else
{
if (k <= (n / 3))
printf("1 %I64d\n", k * 2);
else
{
LL need = n / 3;
LL res = k - need;
printf("1 %I64d\n", need * 2 - (res - (n - need * 3)));
}
}
}
}
return 0;
}
C. Planning
贪心枚举当前时间k，让1-k时间内花费最多的航班最优降落即可，弱比我想到的是线段树，而大牛都是set……
代码：
#include <stdio.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 6e5 + 7;
struct seg
{
int l, mid, r;
int v, id;
} T[N << 2];
int arr[N];
int Time[N];
inline void pushup(int k)
{
if (T[LC(k)].v > T[RC(k)].v)
{
T[k].v = T[LC(k)].v;
T[k].id = T[LC(k)].id;
}
else
{
T[k].v = T[RC(k)].v;
T[k].id = T[RC(k)].id;
}
}
void build(int k, int l, int r)
{
T[k].l = l;
T[k].r = r;
T[k].mid = MID(l, r);
if (l == r)
{
T[k].v = arr[l];
T[k].id = l;
}
else
{
build(LC(k), l, T[k].mid);
build(RC(k), T[k].mid + 1, r);
pushup(k);
}
}
void update(int k, int x)
{
if (T[k].l == T[k].r)
T[k].v = -INF;
else
{
if (x <= T[k].mid)
update(LC(k), x);
else
update(RC(k), x);
pushup(k);
}
}
int query(int k, int l, int r)
{
if (l <= T[k].l && T[k].r <= r)
return T[k].id;
else
{
if (r <= T[k].mid)
return query(LC(k), l, r);
else if (l > T[k].mid)
return query(RC(k), l, r);
else
{
int ll = query(LC(k), l, r);
int rr = query(RC(k), l, r);
if (arr[ll] > arr[rr])
return ll;
else
return rr;
}
}
}
int main(void)
{
int n, k, i;
while (~scanf("%d%d", &n, &k))
{
for (i = 1; i <= n; ++i)
scanf("%d", &arr[i]);
build(1, 1, n);
LL ans = 0;
for (i = k + 1; i <= k + n; ++i)
{
int pos = query(1, 1, min(i, n));
ans = ans + (LL)(arr[pos]) * (LL)(i - pos);
update(1, pos);
arr[pos] = -INF;
Time[pos] = i;
}
printf("%I64d\n", ans);
for (i = 1; i <= n; ++i)
printf("%d%c", Time[i], " \n"[i == n]);
}
return 0;
}
D. Jury Meeting
前缀+后缀思想，主要用两个二维数组pre和suf和维护当前某地最少花费的cost数组，用pre[i][0]表示当前到第i天为止所有能到达0点的人最少花费，pre[i][1]表示有几个人能到达0点，因此枚举i，把航班天数小于等于i的数据都拿去更新cost，再结合cost更新pre，类似于前缀最小值一样处理一遍，然后返程suf数组做一个与pre相反顺序的后缀最小花费处理，然后枚举到达和离开的区间[l,l+k+1]，检查一下区间合法性和区间内是否可以到齐，再更新答案即可
代码；
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1000010;
struct info
{
int d, s, t, c;
};
info A[N], B[N];
LL pre[N][2], suf[N][2], cost[N];
void init()
{
CLR(pre, 0);
CLR(suf, 0);
}
int main(void)
{
int n, m, k, i;
while (~scanf("%d%d%d", &n, &m, &k))
{
init();
int cnta = 0, cntb = 0;
int maxday = 0;
for (i = 0; i < m; ++i)
{
int d, s, t, c;
scanf("%d%d%d%d", &d, &s, &t, &c);
if (t == 0)
A[cnta++] = (info) {d, s, t, c};
else
B[cntb++] = (info) {d, s, t, c};
maxday = max(maxday, d);
}
sort(A, A + cnta, [](info a, info b) {return a.d < b.d;});
sort(B, B + cntb, [](info a, info b) {return a.d > b.d;});
LL ps = 0, pc = 0;
int cur = 0;
CLR(cost, 0);
for (i = 1; i <= maxday; ++i)
{
while (A[cur].d <= i && cur < cnta)
{
if (!cost[A[cur].s])
{
cost[A[cur].s] = A[cur].c;
ps += A[cur].c;
++pc;
}
else if (cost[A[cur].s] > A[cur].c)
{
ps -= cost[A[cur].s];
ps += A[cur].c;
cost[A[cur].s] = A[cur].c;
}
++cur;
}
pre[i][0] = ps;
pre[i][1] = pc;
}
cur = 0;
ps = 0;
pc = 0;
CLR(cost, 0);
for (i = maxday; i >= 1; --i)
{
while (B[cur].d >= i && cur < cntb)
{
if (!cost[B[cur].t])
{
cost[B[cur].t] = B[cur].c;
ps += B[cur].c;
++pc;
}
else if (cost[B[cur].t] > B[cur].c)
{
ps -= cost[B[cur].t];
ps += B[cur].c;
cost[B[cur].t] = B[cur].c;
}
++cur;
}
suf[i][0] = ps;
suf[i][1] = pc;
}
LL ans = 0x3f3f3f3f3f3f3f3f;
for (i = 1; i <= maxday; ++i)
{
if (i + k + 1 <= maxday && pre[i][1] >= n && suf[i + k + 1][1] >= n && pre[i][0] + suf[i + k + 1][0] < ans)
ans = pre[i][0] + suf[i + k + 1][0];
}
printf("%I64d\n", ans == 0x3f3f3f3f3f3f3f3f ? -1 : ans);
}
return 0;
}
