CSP 2021 复盘
作者：互联网
T1：
题目大意：
有 \(n\) 个廊桥，\(m_1\) 个一类飞机、\(m_2\) 个二类飞机，贪心原则分配廊桥，问做多能给多少飞机分上廊桥。
思路：
设 \(f_i\) 表示分 \(i\) 个廊桥给一类飞机的最多的飞机，\(g_i\) 表示分 \(i\) 个廊桥给二类飞机的最多的飞机。题目就转化为求最大的 \(f_i+g_{n-i}\)
考场上想的是三分，但是用脚趾都能造出一个卡三分的数据。
实际上可以发现一个飞机只要在 \(x\) 个廊桥的情况下能分配到，那么 \([x,\infty]\) 都能。先用堆求出每个飞机能分配最少需要的廊桥，然后解法显然。
代码：
const int N = 1e5 + 10;
inline ll Read() {
ll x = 0, f = 1;
char c = getchar();
while (c != '-' && (c < '0' || c > '9')) c = getchar();
if (c == '-') f = -f, c = getchar();
while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
return x * f;
}
int n, m1, m2;
struct node {
int l, r;
bool operator < (node a) const { return l < a.l; }
} a[N], b[N];
int pa[N], pb[N], ca[N], cb[N];
priority_queue<int, vector<int>, greater<int> > q;
struct pQueue {
int val, id;
bool operator < (const pQueue &a) const { return val > a.val; }
};
priority_queue<pQueue> qr;
int suma = 0, sumb = 0, ans;
int main() {
// freopen(".in", "r", stdin);
// freopen(".out", "w", stdout);
n = Read(), m1 = Read(), m2 = Read();
for (int i = 1; i <= m1; i++)
a[i].l = Read(), a[i].r = Read();
for (int i = 1; i <= m2; i++)
b[i].l = Read(), b[i].r = Read();
sort (a + 1, a + 1 + m1);
sort (b + 1, b + 1 + m2);
for (int i = 1; i <= n; i++) q.push(i);
for (int i = 1; i <= m1; i++) {
if (!qr.empty())
for (pQueue j = qr.top(); j.val < a[i].l && !qr.empty(); ) {
qr.pop();
q.push(j.id);
if (!qr.empty()) j = qr.top();
}
if (!q.empty()) {
pa[i] = q.top();
ca[pa[i]]++;
qr.push((pQueue){a[i].r, pa[i]});
q.pop();
}
}
for (; !q.empty(); q.pop());
for (; !qr.empty(); qr.pop());
for (int i = 1; i <= n; i++) q.push(i);
for (int i = 1; i <= m2; i++) {
if (!qr.empty())
for (pQueue j = qr.top(); j.val < b[i].l && !qr.empty(); ) {
qr.pop();
q.push(j.id);
if (!qr.empty()) j = qr.top();
}
if (!q.empty()) {
pb[i] = q.top();
cb[pb[i]]++;
sumb++;
qr.push((pQueue){b[i].r, pb[i]});
q.pop();
}
}
for (int i = 0; i <= n; i++) {
suma += ca[i], sumb -= cb[n - i + 1];
ans = max(ans, suma + sumb);
}
printf ("%d\n", ans);
return 0;
}
T2：
思路：
有一个很强的状态 \(f_{l,r,[0,5]}\) 表示在区间 \([l,r]\) 中字符串属于 全部是*、左右为匹配的括号、左括右*、左右不一定匹配的括号、左*右括，左右为*。
这个状态设法不用判重太强了。
转移见代码。
代码：
const int N = 510;
const ll mod = 1e9 + 7;
inline ll Read() {
ll x = 0, f = 1;
char c = getchar();
while (c != '-' && (c < '0' || c > '9')) c = getchar();
if (c == '-') f = -f, c = getchar();
while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
return x * f;
}
int n, m;
char s[N];
ll f[N][N][10];
int main() {
// freopen(".in", "r", stdin);
// freopen(".out", "w", stdout);
n = Read(), m = Read();
scanf ("%s", s + 1);
for (int i = 1; i <= n; i++) f[i][i - 1][0] = 1;
for (int len = 1; len <= n; len++)
for (int l = 1; l + len - 1 <= n; l++) {
int r = l + len - 1;
if (len <= m) f[l][r][0] = f[l][r - 1][0] * (s[r] == '?' || s[r] == '*');
if (len >= 2) {
if ((s[l] == '(' || s[l] == '?') && (s[r] == ')' || s[r] == '?'))
f[l][r][1] = (f[l + 1][r - 1][0] + f[l + 1][r - 1][2] + f[l + 1][r - 1][3] + f[l + 1][r - 1][4]) % mod;
for (int k = l; k < r; k++)
(f[l][r][2] += f[l][k][3] * f[k + 1][r][0] % mod) %= mod,
(f[l][r][3] += (f[l][k][2] + f[l][k][3]) % mod * f[k + 1][r][1] % mod) %= mod,
(f[l][r][4] += (f[l][k][4] + f[l][k][5]) % mod * f[k + 1][r][1] % mod) %= mod,
(f[l][r][5] += f[l][k][4] * f[k + 1][r][0] % mod) %= mod;
}
f[l][r][3] = (f[l][r][3] + f[l][r][1]) % mod;
f[l][r][5] = (f[l][r][5] + f[l][r][0]) % mod;
}
printf ("%lld\n", f[1][n][3]);
return 0;
}
T3：
思路：
你看它像个栈，那就把它当栈做。没了。
代码：
const int N = 1e6 + 10;
inline ll Read() {
ll x = 0, f = 1;
char c = getchar();
while (c != '-' && (c < '0' || c > '9')) c = getchar();
if (c == '-') f = -f, c = getchar();
while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
return x * f;
}
int t, n, sm, em;
int a[N];
char b[N];
int top1, bot1, top2, bot2;
bool Work(int opt = 1) {
int pos = 0; top1 = top2 = 0, bot1 = bot2 = 1;
if (opt) {
for (int i = 2; i <= 2 * n; i++)
if (a[i] == a[1]) { pos = i; break; }
bot1 = 2, top1 = pos - 1, bot2 = pos + 1, top2 = 2 * n;
} else {
for (int i = 2 * n - 1; i; i--)
if (a[i] == a[2 * n]) { pos = i; break; }
bot1 = 1, top1 = pos - 1, bot2 = pos + 1, top2 = 2 * n - 1;
}
bool flag = 0;
for (int i = 1; i < n; i++) {
if (bot1 <= top1 && (a[top1] == a[bot1] && bot1 < top1 || a[bot1] == a[bot2] && bot2 <= top2)) {
if (bot1 < top1 && a[top1] == a[bot1])
top1--, bot1++, b[++sm] = 'L', b[--em] = 'L';
else bot1++, bot2++, b[++sm] = 'L', b[--em] = 'R';
} else {
if (bot2 <= top2 && (a[top2] == a[top1] && bot1 <= top1 || a[top2] == a[bot2] && bot2 < top2)) {
if (bot2 < top2 && a[top2] == a[bot2])
top2--, bot2++, b[++sm] = 'R', b[--em] = 'R';
else top2--, top1--, b[++sm] = 'R', b[--em] = 'L';
} else return 0;
}
}
printf ("%s\n", b + 1);
return 1;
}
int main() {
// freopen(".in", "r", stdin);
// freopen(".out", "w", stdout);
for (t = Read(); t--; ) {
n = Read();
for (int i = 1; i <= 2 * n; i++) a[i] = Read();
for (int i = 1; i <= 2 * n + 1; i++) b[i] = 0;
b[sm = 1] = b[em = 2 * n] = 'L';
if (!Work()) {
for (int i = 1; i <= 2 * n + 1; i++) b[i] = 0;
b[sm = 1] = 'R', b[em = 2 * n] = 'L';
if (!Work(0)) printf ("-1\n");
}
}
return 0;
}
T4：
咕咕咕。
标签：int,ll,CSP,2021,&&,廊桥,mod,复盘,getchar 来源： https://www.cnblogs.com/GJY-JURUO/p/15570378.html