AIM Tech Round 5 (rated, Div. 1 + Div. 2)
Posted kisekipurin2019
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A - Find Square
没意思。
int n, m;
char g[205][205];
void test_case() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i)
scanf("%s", g[i] + 1);
int l1 = -1, l2 = -1;
int s1 = -1, s2 = -1;
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
if(g[i][j] == 'B') {
l1 = i;
l2 = j;
if(s1 == -1)
s1 = i;
if(s2 == -1)
s2 = j;
}
}
}
printf("%d %d
", (s1 + l1) / 2, (s2 + l2) / 2);
}B - Unnatural Conditions
没意思。
int n, m;
void test_case() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= 2200; ++i)
putchar('5');
putchar('5');
putchar('
');
for(int i = 1; i <= 2200; ++i)
putchar('4');
putchar('5');
putchar('
');
}C - Rectangles
假算法：
int n;
int x1[150005], y1[150005];
int x2[150005], y2[150005];
int xx[300005], yy[300005], top, xtop, ytop;
int dx[300005], dy[300005];
int dx2[300005], dy2[300005];
bool check(int x, int y) {
printf("x=%d y=%d
", x, y);
int LX = dx2[x - 1];
int RX = n - dx[x];
if(LX + RX >= 2)
return false;
int LY = dy2[y - 1];
int RY = n - dy[y];
if(LY + RY >= 2)
return false;
printf("LX=%d RX=%d LY=%d RY=%d
", LX, RX, LY, RY);
return true;
}
void test_case() {
scanf("%d", &n);
top = 0;
for(int i = 1; i <= n; ++i) {
scanf("%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i]);
++top;
xx[top] = x1[i];
yy[top] = y1[i];
++top;
xx[top] = x2[i];
yy[top] = y2[i];
}
sort(xx + 1, xx + 1 + top);
xtop = unique(xx + 1, xx + 1 + top) - (xx + 1);
sort(yy + 1, yy + 1 + top);
ytop = unique(yy + 1, yy + 1 + top) - (yy + 1);
for(int i = 1; i <= n; ++i) {
x1[i] = lower_bound(xx + 1, xx + 1 + xtop, x1[i]) - (xx);
x2[i] = lower_bound(xx + 1, xx + 1 + xtop, x2[i]) - (xx);
y1[i] = lower_bound(yy + 1, yy + 1 + ytop, y1[i]) - (yy);
y2[i] = lower_bound(yy + 1, yy + 1 + ytop, y2[i]) - (yy);
printf("%d %d %d %d
", x1[i], y1[i], x2[i], y2[i]);
}
for(int i = 1; i <= n; ++i) {
++dx[x1[i]];
++dx2[x2[i]];
++dy[y1[i]];
++dy2[y2[i]];
}
for(int i = 1; i <= xtop; ++i) {
dx[i] += dx[i - 1];
dx2[i] += dx2[i - 1];
}
for(int i = 1; i <= ytop; ++i) {
dy[i] += dy[i - 1];
dy2[i] += dy2[i - 1];
}
for(int i = 1; i <= n; ++i) {
if(check(x1[i], y1[i])) {
printf("%d %d
", xx[x1[i]], yy[y1[i]]);
return;
}
if(check(x1[i], y2[i])) {
printf("%d %d
", xx[x1[i]], yy[y2[i]]);
return;
}
if(check(x2[i], y1[i])) {
printf("%d %d
", xx[x2[i]], yy[y1[i]]);
return;
}
if(check(x2[i], y2[i])) {
printf("%d %d
", xx[x2[i]], yy[y2[i]]);
return;
}
}
}原因：x和y是不能分开考虑的。过不了第一个样例。
正确做法：扫描线。
int n;
int x1[150005], y1[150005];
int x2[150005], y2[150005];
int xx[300005], yy[300005], top, xtop, ytop;
vector<pii> upd[300005];
struct SegmentTree {
#define ls (o<<1)
#define rs (o<<1|1)
static const int MAXN = 300000;
static const int INF = 0x3f3f3f3f;
int ma[(MAXN << 2) + 5];
int ps[(MAXN << 2) + 5];
int lz[(MAXN << 2) + 5];
void PushUp(int o) {
if(ma[ls] >= ma[rs]) {
ma[o] = ma[ls];
ps[o] = ps[ls];
} else {
ma[o] = ma[rs];
ps[o] = ps[rs];
}
}
void PushDown(int o, int l, int r) {
if(lz[o]) {
lz[ls] += lz[o];
lz[rs] += lz[o];
ma[ls] += lz[o];
ma[rs] += lz[o];
lz[o] = 0;
}
}
void Build(int o, int l, int r) {
if(l == r) {
ma[o] = 0;
ps[o] = l;
} else {
int m = l + r >> 1;
Build(ls, l, m);
Build(rs, m + 1, r);
PushUp(o);
}
lz[o] = 0;
}
void Update(int o, int l, int r, int ql, int qr, int v) {
if(ql <= l && r <= qr) {
lz[o] += v;
ma[o] += v;
} else {
PushDown(o, l, r);
int m = l + r >> 1;
if(ql <= m)
Update(ls, l, m, ql, qr, v);
if(qr >= m + 1)
Update(rs, m + 1, r, ql, qr, v);
PushUp(o);
}
}
pii QueryMax(int o, int l, int r, int ql, int qr) {
if(ql <= l && r <= qr) {
return {ma[o], ps[o]};
} else {
PushDown(o, l, r);
int m = l + r >> 1;
pii res = {-INF, -1};
if(ql <= m)
res = QueryMax(ls, l, m, ql, qr);
if(qr >= m + 1) {
pii tmp = QueryMax(rs, m + 1, r, ql, qr);
if(tmp.first > res.first)
res = tmp;
}
return res;
}
}
#undef ls
#undef rs
} st;
void test_case() {
scanf("%d", &n);
top = 0;
for(int i = 1; i <= n; ++i) {
scanf("%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i]);
++y2[i];
++top;
xx[top] = x1[i];
yy[top] = y1[i];
++top;
xx[top] = x2[i];
yy[top] = y2[i];
}
sort(xx + 1, xx + 1 + top);
xtop = unique(xx + 1, xx + 1 + top) - (xx + 1);
sort(yy + 1, yy + 1 + top);
ytop = unique(yy + 1, yy + 1 + top) - (yy + 1);
for(int i = 1; i <= n; ++i) {
x1[i] = lower_bound(xx + 1, xx + 1 + xtop, x1[i]) - (xx);
x2[i] = lower_bound(xx + 1, xx + 1 + xtop, x2[i]) - (xx);
y1[i] = lower_bound(yy + 1, yy + 1 + ytop, y1[i]) - (yy);
y2[i] = lower_bound(yy + 1, yy + 1 + ytop, y2[i]) - (yy);
//printf("%d %d %d %d
", x1[i], y1[i], x2[i], y2[i]);
upd[y1[i]].push_back({x1[i], x2[i]});
upd[y2[i]].push_back({-x1[i], -x2[i]});
}
st.Build(1, 1, xtop);
for(int i = 1; i <= ytop; ++i) {
for(int j = 0; j < upd[i].size(); ++j) {
int l = upd[i][j].first;
int r = upd[i][j].second;
if(l > 0 && r > 0)
st.Update(1, 1, xtop, l, r, 1);
else
st.Update(1, 1, xtop, -l, -r, -1);
}
pii res = st.QueryMax(1, 1, xtop, 1, xtop);
if(res.first >= n - 1) {
int X = res.second;
int Y = i;
printf("%d %d
", xx[X], yy[Y]);
return;
}
}
printf("-1 -1
");
exit(-1);
}以上是关于AIM Tech Round 5 (rated, Div. 1 + Div. 2)的主要内容，如果未能解决你的问题，请参考以下文章
AIM Tech Round 5 (rated, Div. 1 + Div. 2) BUnnatural Conditions
AIM Tech Round 5 (rated, Div. 1 + Div. 2) CRectangles
AIM Tech Round 5 (rated, Div. 1 + Div. 2) A Find Square