JOISC2020 题解
Day1T1 建筑装饰4
题目链接：Day1T1 建筑装饰4
Solution
我们先考虑朴素的\(dp\)方法：
设\(dp_{i,j,k}\)表示前\(i\)个数中，选了\(j\)个\(B\)数组中的数，并且第\(i\)个数选的是\(A/B\)时，是否存在满足单调不降的数列。
复杂度：\(O(n^2)\)，期望得分：\(11\)分。
考虑如何优化。可以打个表，发现当\(i,k\)固定的时候，满足\(dp_{i,j,k}=1\)的\(j\)刚好是一个区间。
我们可以大胆猜结论：如果\(dp_{i,j_1,k}=1\)并且\(dp_{i,j_2,k}=1\)，那么对于任意\(j_1\le j\le j_2\)，都满足\(dp_{i,j,k}=1\)。
于是，我们可以记录数组\(l_{i,k}\)和\(r_{i,k}\)分别表示最小的\(j\)和最大的\(j\)。
输出路径的时候从后往前找即可。
复杂度：\(O(n)\)，期望得分：\(100\)分。
Code
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int inf = 1e9;
const int N = 1000005;
int l[N][2], r[N][2];
int a[N][2], n;
void ckmin(int &a, int b) {
if (a > b) a = b;
}
void ckmax(int &a, int b) {
if (a < b) a = b;
}
int main() {
// freopen("building.in", "r", stdin);
// freopen("building.out", "w", stdout);
n = read();
a[0][0] = a[0][1] = -inf, a[2 * n + 1][0] = a[2 * n + 1][1] = inf;
for (rint i = 1; i <= 2 * n; i++) a[i][0] = read();
for (rint i = 1; i <= 2 * n; i++) a[i][1] = read();
for (rint i = 1; i <= 2 * n; i++) {
for (rint j = 0; j <= 1; j++) {
l[i][j] = inf, r[i][j] = -inf;
for (rint k = 0; k <= 1; k++) {
if (a[i][j] >= a[i - 1][k]) {
ckmin(l[i][j], l[i - 1][k] + (j == 0));
ckmax(r[i][j], r[i - 1][k] + (j == 0));
}
}
}
}
stack <int> ans;
if (l[2 * n][0] <= n && n <= r[2 * n][0]) {
int lef = n, st = 0;
ans.push(0);
for (rint i = 2 * n - 1; i >= 1; i--) {
if (st == 0) {
lef--;
for (rint j = 0; j <= 1; j++) {
if (a[i + 1][st] >= a[i][j] && l[i][j] <= lef && lef <= r[i][j]) {
st = j;
break;
}
}
} else {
for (rint j = 0; j <= 1; j++) {
if (a[i + 1][st] >= a[i][j] && l[i][j] <= lef && lef <= r[i][j]) {
st = j;
break;
}
}
}
ans.push(st);
}
while (!ans.empty()) {
if (ans.top() == 0) putchar('A');
else putchar('B');
ans.pop();
}
puts("");
} else if (l[2 * n][1] <= n && n <= r[2 * n][1]) {
int lef = n, st = 1;
ans.push(1);
for (rint i = 2 * n - 1; i >= 1; i--) {
if (st == 0) {
lef--;
for (rint j = 0; j <= 1; j++) {
if (a[i + 1][st] >= a[i][j] && l[i][j] <= lef && lef <= r[i][j]) {
st = j;
break;
}
}
} else {
for (rint j = 0; j <= 1; j++) {
if (a[i + 1][st] >= a[i][j] && l[i][j] <= lef && lef <= r[i][j]) {
st = j;
break;
}
}
}
ans.push(st);
}
while (!ans.empty()) {
if (ans.top() == 0) putchar('A');
else putchar('B');
ans.pop();
}
puts("");
} else {
puts("-1");
}
return 0;
}
Day1T2 汉堡肉
题目链接：Day1T2 汉堡肉
Solution
对于\(k=1\)，显然这个点位于这\(n\)个矩形的交。
复杂度：\(O(n)\)，期望得分：\(2\)分。
对于\(k=2\)，令\(r=min_{i=1}^{n}R_i\)，令其中一个点的横坐标为\(r\)，同时枚举它所有的可能纵坐标，并且包含这个点的矩形删掉。
剩余的矩形按\(k=1\)的做法即可得到另一个点。
复杂度：\(O(n^2)\)，期望得分：\(1\)分，结合\(k=1\)可得\(3\)分。
考虑如何优化。我们先将所有的矩形按照\(R_i\)优化，然后在扫描线的过程中记录剩余矩形中的\(max(L_i),max(D_i),min(R_i),min(U_i)\)。
具体过程可以用优先队列来实现。
复杂度：\(O(nlogn)\)，期望得分：\(3\)分，结合\(k=1\)可得\(6\)分。
对于剩下的点，我采用了随机化做法。
先随机打乱这\(n\)个矩形，然后将前\(k\)个矩形分别放置这\(k\)类。
接下来，将第\(k+1到n\)这些矩形依次插入这\(k\)类。
定义估价函数\(h(rec_x,rec_y)=\frac{area(merge(rec_x,rec_y))}{area(rec_y)}\)表示覆盖率。
我们可以去寻找第\(i\)个矩形与第\(j\)类矩阵交后的估价函数最大值。
那么估价函数越高，即越优秀，匹配成功的概率最大。
按照如上方式随机化，可以通过此题。
复杂度：\(O(欧皇)\)，期望得分：\(100\)分。
Code
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int N = 200005;
int n, k;
struct rect {
ll x1, y1, x2, y2;
rect(ll _x1 = 0, ll _y1 = 0, ll _x2 = 0, ll _y2 = 0) {
x1 = _x1, y1 = _y1, x2 = _x2, y2 = _y2;
}
} a[N], b[5];
rect merge(rect x, rect y) {
return rect(max(x.x1, y.x1), max(x.y1, y.y1), min(x.x2, y.x2), min(x.y2, y.y2));
}
ll area(rect x) { // 矩形内包含多少点
if (x.x1 > x.x2 || x.y1 > x.y2) return 0;
else return (x.x2 - x.x1 + 1) * (x.y2 - x.y1 + 1);
}
double h(rect x, rect y) {
rect m = merge(x, y);
return area(m) / 1.0 / area(y);
}
int main() {
srand(time(NULL));
scanf("%d%d", &n, &k);
for (rint i = 1; i <= n; i++) {
scanf("%lld%lld%lld%lld", &a[i].x1, &a[i].y1, &a[i].x2, &a[i].y2);
}
while (true) {
int ok = 1;
random_shuffle(a + 1, a + n + 1);
for (rint i = 1; i <= k; i++) b[i] = a[i];
for (rint i = k + 1; i <= n; i++) {
double max_h = -1e9; int id = 1;
for (rint j = 1; j <= k; j++) {
double _h = h(a[i], b[j]);
if (_h > max_h) {
max_h = _h;
id = j;
}
}
b[id] = merge(b[id], a[i]);
if (area(b[id]) <= 0) {
ok = 0;
break;
}
}
if (ok) break;
}
for (rint i = 1; i <= k; i++) {
printf("%lld %lld\n", b[i].x1, b[i].y1);
}
return 0;
}
谈谈正经的做法。
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define all(x) x.begin(), x.end()
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
inline void ckmin(int &a, int b) {
if (a > b) a = b;
}
inline void ckmax(int &a, int b) {
if (a < b) a = b;
}
const int N = 1500005;
const int inf = 1e9;
struct rect {
int x1, y1, x2, y2;
} a[N];
int n, K;
pair <int, int> ans[5];
int cnt[N];
void modify(int x, int y, int v) {
for (rint i = 1; i <= n; i++) {
if (a[i].x1 <= x && x <= a[i].x2 && a[i].y1 <= y && y <= a[i].y2) {
cnt[i] += v;
}
}
}
void dfs(int d) {
int x1 = -inf, x2 = inf, y1 = -inf, y2 = inf;
for (rint i = 1; i <= n; i++) if (!cnt[i]) {
ckmax(x1, a[i].x1), ckmax(y1, a[i].y1);
ckmin(x2, a[i].x2), ckmin(y2, a[i].y2);
}
if (d == 1) {
if (x1 <= x2 && y1 <= y2) { // 最后一个点在剩余矩形的交内
ans[1] = make_pair(x1, y1);
for (rint i = 1; i <= K; i++) {
printf("%d %d\n", ans[i].first, ans[i].second);
}
exit(0);
}
return ;
}
for (rint cx = 0; cx < 2; cx++) {
for (rint cy = 0; cy < 2; cy++) {
int px = (cx ? x1 : x2);
int py = (cy ? y1 : y2);
modify(px, py, 1);
ans[d] = make_pair(px, py);
dfs(d - 1);
modify(px, py, -1);
}
}
}
vector <int> h[N];
vector <tuple <int, int, int>> sl[4], sr[4];
int pre[4][N], suf[4][N], id[N][4];
int fir(tuple <int, int, int> a) {
return get<0>(a);
}
int sec(tuple <int, int, int> b) {
return get<1>(b);
}
int thi(tuple <int, int, int> c) {
return get<2>(c);
}
vector <int> G[N];
void add(int u, int v) { G[u].push_back(v); }
int dfn[N], low[N], dtot;
int ins[N], be[N], scc;
stack <int> st;
void tarjan(int u) {
dfn[u] = low[u] = ++dtot;
st.push(u), ins[u] = 1;
for (auto v: G[u]) {
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (ins[v]) {
low[u] = min(low[u], dfn[v]);
}
}
if (dfn[u] == low[u]) {
int y; ++scc;
while (y = st.top()) {
ins[y] = 0, st.pop();
be[y] = scc;
if (y == u) break;
}
}
}
void solve() {
int x1 = -inf, x2 = inf, y1 = -inf, y2 = inf;
for (rint i = 1; i <= n; i++) {
ckmax(x1, a[i].x1), ckmax(y1, a[i].y1);
ckmin(x2, a[i].x2), ckmin(y2, a[i].y2);
}
//cerr << "ok1 !\n";
//assert(x1 >= x2 && y1 >= y2);
int tot = 0, min_h = inf;
for (rint i = 1; i <= n; i++) {
vector <int> o = h[i];
// 两条纵边 (0 <= id <= 1)
if (a[i].x1 <= x1 && x1 <= a[i].x2) o.pb(0);
if (a[i].x1 <= x2 && x2 <= a[i].x2) o.pb(1);
// 两条横边 (2 <= id <= 3)
if (a[i].y1 <= y1 && y1 <= a[i].y2) o.pb(2);
if (a[i].y1 <= y2 && y2 <= a[i].y2) o.pb(3);
ckmin(min_h, o.size());
if (o.size() >= 3) { // 一定与一条边有交
//o.clear();
continue;
}
id[i][0] = ++tot, id[i][1] = ++tot;
if (o.size() == 1) add(id[i][1], id[i][0]);
int siz = o.size();
for (rint j = 0; j < siz; j++) {
sl[o[j]].emplace_back(o[j] < 2 ? max(y2, a[i].y1) : max(x2, a[i].x1), id[i][j], id[i][j ^ 1]);
sr[o[j]].emplace_back(o[j] < 2 ? min(y1, a[i].y2) : min(x1, a[i].x2), id[i][j], id[i][j ^ 1]);
}
//o.pb(0);
h[i] = o;
//printf("%d\n", h[i].size());
}
//exit(0);
//cerr << "min_h" << ' ' << min_h << '\n';
//assert(min_h > 0);
//cerr << "ok2 !\n";
for (rint v = 0; v < 4; v++) {
sort(all(sl[v])), sort(all(sr[v]));
int sz_l = sl[v].size(), sz_r = sr[v].size();
for (rint i = 0; i < sz_r; i++) {
pre[v][i + 1] = ++tot;
if (i > 0) add(pre[v][i + 1], pre[v][i]);
add(pre[v][i + 1], thi(sr[v][i]));
}
for (rint i = 0, j = 0; i < sz_l; i++) {
while (j < sz_r && fir(sr[v][j]) < fir(sl[v][i])) j++;
if (j > 0) add(sec(sl[v][i]), pre[v][j]);
}
for (rint i = sz_l - 1; i >= 0; i--) {
suf[v][i] = ++tot;
if (i + 1 < sz_l) add(suf[v][i], suf[v][i + 1]);
add(suf[v][i], thi(sl[v][i]));
}
for (rint i = sz_r - 1, j = sz_l; i >= 0; i--) {
while (j > 0 && fir(sl[v][j - 1]) > fir(sr[v][i])) j--;
if (j < sz_l) add(sec(sr[v][i]), suf[v][j]);
}
}
//printf("tot=%d\n", tot);
//for (rint i = 1; i <= tot; i++) {
// printf("%d\n", G[i].size());
//}
//exit(0);
//cerr << "ok3 !\n";
for (rint i = 1; i <= tot; i++) if (!dfn[i]) {
tarjan(i);
}
//cerr << "ok4 !\n";
vector <int> L(5), R(4);
L[0] = L[1] = y2, L[2] = L[3] = x2;
R[0] = R[1] = y1, R[2] = R[3] = x1;
//cerr << "ok5 !\n";
for (rint i = 1; i <= n; i++) {
if (!id[i][0]) continue;
assert(h[i].size());
int j = h[i][be[id[i][0]] >= be[id[i][1]]]; // 拓扑序反过来
//printf("%d: %d\n", i, be[id[i][0]] >= be[id[i][1]]);
ckmax(L[j], j < 2 ? a[i].y1 : a[i].x1);
ckmin(R[j], j < 2 ? a[i].y2 : a[i].x2);
}
//cerr << "ok6! \n";
assert(L[0] <= R[0] && L[1] <= R[1] && L[2] <= R[2] && L[3] <= R[3]);
printf("%d %d\n", x2, L[1]);
printf("%d %d\n", x1, L[0]);
printf("%d %d\n", L[3], y2);
printf("%d %d\n", L[2], y1);
}
int main() {
//freopen("04-14.txt.in", "r", stdin);
//freopen("my.out", "w", stdout);
scanf("%d%d", &n, &K);
for (rint i = 1; i <= n; i++) {
scanf("%d%d%d%d", &a[i].x1, &a[i].y1, &a[i].x2, &a[i].y2);
}
dfs(K);
assert(K == 4);
solve();
return 0;
}
Day1T3 扫除
题目链接：Day1T3 扫除
Solution
Code
咕咕咕
Day2T1 变色龙之恋
题目链接：Day2T1 变色龙之恋
Day2T2 有趣的Joitter交友
题目链接：Day2T2 有趣的Joitter交友
Day2T3 遗迹
题目链接：Day2T3 遗迹
Day3T1 星座3
题目链接：Day3T1 星座3
Day3T2 收获
题目链接：Day3T2 收获
Day3T3 迷路的猫
题目链接：Day3T3 迷路的猫
Day4T1 首都
题目链接：Day4T1 首都
Day4T2 制作团子
题目链接：Day4T2 制作团子
Day4T3 应对方案
题目链接：Day4T3 应对方案
