Codeforces Round #344 (Div. 2)
注意是或不是异或
#include <bits/stdc++.h>
int a[1005], b[1005];
int main() {
int n; scanf ("%d", &n);
for (int i=0; i<n; ++i) {
scanf ("%d", a+i);
}
for (int i=0; i<n; ++i) {
scanf ("%d", b+i);
}
int ans = 0;
for (int i=0; i<n; ++i) {
int x = 0, y = 0;
for (int j=i; j<n; ++j) {
x |= a[j]; y |= b[j];
ans = std::max (ans, x + y);
}
}
printf ("%d\n", ans);
return 0;
}
color[i][j] = max (timei(rowi), timej(colj)))
#include <bits/stdc++.h>
const int N = 5e3 + 5;
int a[N][N];
std::pair<int, int> row[N], col[N];
int main() {
int n, m, k; scanf ("%d%d%d", &n, &m, &k);
for (int op, x, y, i=1; i<=k; ++i) {
scanf ("%d%d%d", &op, &x, &y);
if (op == 1) {
row[x] = std::make_pair (y, i);
} else {
col[x] = std::make_pair (y, i);
}
}
for (int i=1; i<=n; ++i) {
for (int j=1; j<=m; ++j) {
int tr = row[i].second, tc = col[j].second;
if (tr > tc && tr > 0) {
a[i][j] = row[i].first;
} else if (tc > tr && tc > 0) {
a[i][j] = col[j].first;
}
}
}
for (int i=1; i<=n; ++i) {
for (int j=1; j<=m; ++j) {
printf ("%d%c", a[i][j], j == m ? '\n' : ' ');
}
}
return 0;
}
单调队列+排序 C - Report
题意：给m次操作，每次操作使得[1, r]范围的的a[i]升序或降序排序，问a[i]最后的结果
分析：因为左端点固定，右端点远的且操作时间靠后的可以覆盖之前右端点近的操作，所以先从右端点最远且时间靠后的时间出发，之后类似只要处理右端点降序的位置，即维护一个单调队列。队列中后一次操作长度不大于前一次长度，减少的长度k里的数字为前一次范围中前k大的数或前k小的数。
#include <bits/stdc++.h>
const int N = 2e5 + 5;
int a[N], b[N];
int r[N], t[N];
int main() {
int n, m; scanf ("%d%d", &n, &m);
for (int i=1; i<=n; ++i) {
scanf ("%d", a+i);
}
int s = 0;
for (int x, y, i=1; i<=m; ++i) {
scanf ("%d%d", &x, &y);
while (s > 0 && y >= r[s-1]) s--;
t[s] = x; r[s] = y; s++;
}
r[s++] = 0;
int bl = 1, br = r[0];
for (int i=bl; i<=br; ++i) b[i] = a[i];
std::sort (b+1, b+1+br);
for (int i=1; i<s; ++i) {
for (int j=r[i-1]; j>r[i]; --j) {
a[j] = (t[i-1] == 1) ? b[br--] : b[bl++];
}
}
for (int i=1; i<=n; ++i) {
printf ("%d ", a[i]);
}
puts ("");
return 0;
}
KMP D - Messenger
题意：给两个字符串，问后者在前者里出现的次数。给出的方式：1-a 2-b 3-c 4-d(abbcccddd)
分析：其实想明白就知道这就是简单的KMP问题，文本串第一个和最后一个的个数一定要比模式串多，且里面的要完全相等。还要考虑特殊情况，压缩后m==1或2时。
#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e5 + 5;
struct Part {
ll len; char ch;
bool operator == (const Part &rhs) const {
return len == rhs.len && ch == rhs.ch;
}
bool operator >= (const Part &rhs) const {
return len >= rhs.len && ch == rhs.ch;
}
bool operator != (const Part &rhs) const {
return len != rhs.len || ch != rhs.ch;
}
};
Part A[N], B[N];
int fail[N];
int compress(Part *C, int n) {
int m = 0;
for (int i=1; i<n; ++i) {
if (C[m].ch == C[i].ch) {
C[m].len += C[i].len;
} else {
C[++m] = C[i];
}
}
return m + 1;
}
void get_fail(Part *P, int lenp) {
int i = 0, j = -1; fail[0] = -1;
while (i < lenp) {
if (j == -1 || P[j] == P[i]) {
++i; ++j; fail[i] = j;
} else {
j = fail[j];
}
}
}
ll KMP(Part *C, int n, Part *D, int m) {
get_fail (D, m);
int i = 0, j = 0;
ll ret = 0;
while (i < n) {
while (j != -1 && C[i] != D[j]) j = fail[j];
i++; j++;
if (j == m) {
if (C[i] >= D[m] && C[i-m-1] >= D[-1]) ret++;
j = fail[j];
}
}
return ret;
}
int main() {
int n, m; scanf ("%d%d", &n, &m);
int bug = 0;
char str[3];
for (int i=0; i<n; ++i) {
scanf ("%I64d-%s", &A[i].len, str);
A[i].ch = str[0];
}
for (int i=0; i<m; ++i) {
scanf ("%I64d-%s", &B[i].len, str);
B[i].ch = str[0];
}
n = compress (A, n);
m = compress (B, m);
ll ans = 0;
if (m == 1) {
for (int i=0; i<n; ++i) {
if (A[i].ch == B[0].ch) {
ans += std::max (0ll, A[i].len - B[0].len + 1);
}
}
} else if (m == 2) {
for (int i=0; i<n-1; ++i) {
if (A[i] >= B[0] && A[i+1] >= B[1]) ans++;
}
} else {
ans = KMP (A+1, n-2, B+1, m-2);
}
printf ("%I64d\n", ans);
return 0;
}
题意：将一个点移动到另一个点(可相同)，求
分析：推算化简的得到：
或者
先考虑后者，再次化简得到：，
，可以抽象成ax+b, (x = ar),所以我们枚举ar，得到最大的
，用到了凸包，把直线加入凸包，得到单调上升的类似于
这样的，然后二分找最大值。前者类似。
第一次碰到这种题，写详细点:)
#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e5 + 5;
int a[N];
ll sum[N];
struct Line {
int a; ll b;
ll get(int x) {
return 1ll * a * x + b;
}
};
struct Convex_hull {
int sz;
Line *hull;
Convex_hull(int maxn) {
hull = new Line[++maxn]; sz = 0;
}
void clear() {
sz = 0;
}
bool is_bad(int cur, int pre, int nex) {
Line c = hull[cur], p = hull[pre], n = hull[nex];
return (c.b - n.b) * (c.a - p.a) <= (p.b - c.b) * (n.a - c.a);
}
void add_line(int a, ll b) {
hull[sz++] = (Line) {a, b};
while (sz > 2 && is_bad (sz-2, sz-3, sz-1)) {
hull[sz-2] = hull[sz-1]; sz--;
}
}
ll query(int x) {
int low = -1, high = sz - 1;
while (low + 1 < high) {
int mid = low + high >> 1;
if (hull[mid].get (x) <= hull[mid+1].get (x)) {
low = mid;
} else {
high = mid;
}
}
return hull[high].get (x);
}
};
int main() {
int n; scanf ("%d", &n);
ll ans = 0; sum[0] = 0;
for (int i=1; i<=n; ++i) {
scanf ("%d", a+i);
sum[i] = sum[i-1] + a[i];
ans += 1ll * a[i] * i;
}
Convex_hull *hull = new Convex_hull (n);
ll add = 0;
for (int r=2; r<=n; ++r) {
hull->add_line (r-1, -sum[r-2]);
add = std::max (add, hull->query (a[r]) + sum[r-1] - 1ll * a[r] * r);
}
hull->clear ();
for (int l=n-1; l>=1; --l) {
hull->add_line (-(l+1), -sum[l+1]);
add = std::max (add, hull->query (-a[l]) + sum[l] - 1ll * a[l] * l);
}
printf ("%I64d\n", ans + add);
return 0;
}
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