A. Beru-taxi
就是问,我们知道一个点,从其他点到它的最少花费的时间是多少?
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 7;
double sx, sy;
double ans = 1000000000.;
double x, y, v;
double dis(double x1, double y1, double x2, double y2)
{
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int main()
{
scanf("%lf%lf", &sx, &sy);
int N; scanf("%d", &N);
for(int i=1; i<=N; i++)
{
scanf("%lf%lf%lf", &x, &y, &v);
ans = min(ans, dis(sx, sy, x, y) / v);
}
printf("%lf\n", ans);
return 0;
}
B. Interesting drink
问小于等于某个数的数有多少个,我的做法是O(N)的差分,树状数组等等都是可以搞的,差分显得快一点(因为值的大小小的缘故吧)。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, a[maxN], q, mi, num[maxN];
int main()
{
scanf("%d", &N);
for(int i=1; i<=N; i++)
{
scanf("%d", &a[i]);
num[a[i]]++;
}
for(int i=1; i<maxN; i++) num[i] += num[i - 1];
scanf("%d", &q);
while(q--)
{
scanf("%d", &mi);
if(mi >= maxN) printf("%d\n", N);
else printf("%d\n", num[mi]);
}
return 0;
}
C. Hard problem
简单的DP,但是别忘了条件是等于也算是合法的。
也就是"a == a"也是认为是可行的。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N;
ll c[maxN];
string s[maxN], t[maxN];
struct DO_it
{
ll dp[maxN][2];
inline void init()
{
for(int i=1; i<=N; i++) dp[i][0] = dp[i][1] = INF;
dp[1][0] = 0;
dp[1][1] = c[1];
}
void solve()
{
init();
for(int i=2; i<=N; i++)
{
if(s[i] >= s[i-1]) dp[i][0] = dp[i-1][0];
if(s[i] >= t[i-1]) dp[i][0] = min(dp[i][0], dp[i-1][1]);
if(t[i] >= s[i-1]) dp[i][1] = dp[i-1][0] + c[i];
if(t[i] >= t[i-1]) dp[i][1] = min(dp[i][1], dp[i-1][1] + c[i]);
}
if(dp[N][0] >= INF && dp[N][1] >= INF) printf("-1\n");
else printf("%lld\n", min(dp[N][0], dp[N][1]));
}
}kk;
int main()
{
scanf("%d", &N);
for(int i=1; i<=N; i++) scanf("%lld", &c[i]);
for(int i=1; i<=N; i++)
{
cin >> s[i];
t[i] = s[i];
reverse(t[i].begin(), t[i].end());
}
kk.solve();
return 0;
}
D. Vasiliy's Multiset
简单的字典树了。
题意:
有如下三种操作:
- "+ x"向堆中加入值为x的值;
- "- x"从堆中减去一个值为x的元素,保证x在堆中;
- "? x"询问堆A中与x异或值最大的数的值是多少,输出最大值。
然后就是字典树的一般操作了,可惜卡C卡了点时间,赛后十分钟才写完。
记住一种特殊情况,如果你没往堆中放0,堆中也是有0的。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 8e6 + 7;
int Q;
int x;
char op[3];
int tot = 0, waste[maxN], inwast = 0;
struct node
{
int nex[2], num;
node(int a=0, int b=0, int c=0):nex{a, b}, num(c) {}
void init() { nex[0] = nex[1] = 0; num = 0; }
}trie[maxN];
void Insert(int x)
{
int root = 0;
for(int i=31, id; i>=0; i--)
{
id = (x >> i) & 1;
if(!trie[root].nex[id])
{
if(inwast) trie[root].nex[id] = waste[inwast--];
else trie[root].nex[id] = ++tot;
trie[trie[root].nex[id]].init();
}
root = trie[root].nex[id];
trie[root].num++;
}
}
void del(int x)
{
int root = 0, nex;
for(int i=31, id; i>=0; i--)
{
id = (x >> i) & 1;
nex = trie[root].nex[id];
if(--trie[nex].num == 0)
{
trie[root].nex[id] = 0;
waste[++inwast] = nex;
}
root = nex;
}
}
int fid(int x)
{
int root = 0, ans = 0;
for(int i=31, id; i>=0; i--)
{
id = (x >> i) & 1;
if(trie[root].nex[!id])
{
ans |= (1 << i);
root = trie[root].nex[!id];
}
else root = trie[root].nex[id];
}
return ans;
}
int main()
{
scanf("%d", &Q);
Insert(0);
while(Q--)
{
scanf("%s%d", op, &x);
if(op[0] == '+')
{
Insert(x);
}
else if(op[0] == '-')
{
del(x);
}
else
{
printf("%d\n", fid(x));
}
}
return 0;
}
E. Working routine
可以说是一道模拟题了,写了有些时候,可能是太菜了吧……
题意:给出一个原始矩阵,之后有Q次操作,我们将两个矩阵交换位置,题目中保证两个矩阵不相交,给出的是两个矩阵的左上方的端点,以及它们对应的高和宽。
思路:很明显,题目中只有最多1e4次操作,矩阵的边最大也才不过是1e3,所以很明显的,我们可以O(1e4 * 1e3 * 常数)的“暴力”来解决问题了。
首先,初始化二维链表,我们定义用新的一圈来包围这个子矩阵,相当于是防止溢出了,并且还保护了链首。
然后的话,我们就是要去避免冲突了,我们的更新过程一定是需要有目的的,不然会造成冲突,譬如说这样的一个样例:
3 3 4
1 2 3
4 5 6
7 8 9
3 1 3 3 1 1
3 2 1 3 1 1
2 1 3 2 1 1
3 2 1 1 1 1
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 7;
int N, M, Q, v[maxN][maxN];
struct List_Table
{
int down, right, val;
List_Table(int b=0, int d=0, int f=0):down(b), right(d), val(f) {}
}t[maxN * maxN];
inline int _id(int x, int y) { return x * (M + 2) + y; }
inline void init()
{
for(int i=1; i<=N; i++)
{
for(int j=1; j<=M; j++)
{
t[_id(i, j)] = List_Table(_id(i + 1, j), _id(i, j + 1), v[i][j]);
}
}
for(int i=1; i<=M; i++)
{
t[_id(0, i)] = List_Table(_id(1, i), _id(0, i + 1));
}
t[_id(0, 0)] = List_Table(_id(1, 0), _id(0, 1));
t[_id(0, M + 1)] = List_Table(_id(1, M + 1), 0);
for(int i=1; i<=N; i++)
{
t[_id(i, 0)] = List_Table(_id(i + 1, 0), _id(i, 1));
}
t[_id(N + 1, 0)] = List_Table(0, _id(N + 1, 1));
for(int i=1; i<=N; i++)
{
t[_id(i, M + 1)] = List_Table(_id(i + 1, M + 1), 0);
}
t[_id(N + 1, M + 1)] = List_Table(0, 0);
for(int i=1; i<=M; i++)
{
t[_id(N + 1, i)] = List_Table(0, _id(N + 1, i + 1));
}
}
inline void ex_change(int sx, int sy, int ex, int ey, int h, int w)
{
int id_1 = _id(sx - 1, 0), id_2 = _id(ex - 1, 0), tims = sy - 1, tmp_1, tmp_2;
while(tims --) id_1 = t[id_1].right;
tims = ey - 1;
while(tims --) id_2 = t[id_2].right;
tmp_1 = id_1; tmp_2 = id_2;
for(int i=1; i<=h; i++)
{
id_1 = t[id_1].down; id_2 = t[id_2].down;
swap(t[id_1].right, t[id_2].right);
}
for(int i=1; i<=w; i++)
{
id_1 = t[id_1].right; id_2 = t[id_2].right;
swap(t[id_1].down, t[id_2].down);
}
for(int i=1; i<=w; i++)
{
tmp_1 = t[tmp_1].right; tmp_2 = t[tmp_2].right;
swap(t[tmp_1].down, t[tmp_2].down);
}
for(int i=1; i<=h; i++)
{
tmp_1 = t[tmp_1].down; tmp_2 = t[tmp_2].down;
swap(t[tmp_1].right, t[tmp_2].right);
}
}
inline void Prit()
{
for(int i=1, now, cnt; i<=N; i++)
{
now = t[_id(i, 0)].right;
cnt = M;
while(cnt--)
{
printf("%d ", t[now].val);
now = t[now].right;
}
printf("\n");
}
}
int main()
{
scanf("%d%d%d", &N, &M, &Q);
for(int i=1; i<=N; i++)
{
for(int j=1; j<=M; j++)
{
scanf("%d", &v[i][j]);
}
}
init();
int sx, sy, ex, ey, h, w;
while(Q--)
{
scanf("%d%d%d%d%d%d", &sx, &sy, &ex, &ey, &h, &w);
ex_change(sx, sy, ex, ey, h, w);
// Prit();
}
Prit();
return 0;
}