CSP-J/S2019 做题练习（day5）
A - 中位数图
题解
先找出题意中的\(b\)所在的位置。
再以这个位置为中心，向右\(for\)一遍有多少个大于/小于该数的数
大于就\(++cs\) 小于就\(--cs\)。
因为这个数是中位数，所以大于它的数小于它的数的数量是一样的。
然后每次以\(cs\)为下标的数\(++\)。
因为\(cs\)可能小于零，
所以需要将\(cs\)加上\(100000\)后再\(++\)。
统计答案时直接加上下标为\(0-cs+100000\)的数的值即可。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int n, b, a[100003], p[100003], c, s[100003], cs, cj, sum, ans, pp[200003];
signed main()
{
//File("A");
n = gi(), b = gi();
for (itn i = 1; i <= n; i+=1)
{
a[i] = gi();
if (a[i] == b) c = i;
}
for (int i = c; i <= n; i+=1)
{
if (a[i] > b) ++cs;
if (a[i] < b) --cs;
++pp[cs + 100000];
}
for (int i = c; i >= 1; i-=1)
{
if (a[i] > b) ++cj;
if (a[i] < b) --cj;
ans = ans + pp[0 - cj + 100000];
}
printf("%lld\n", ans);
return 0;
}
B - 树
题解
直接暴力\(dfs\)即可\(AC\)。
数据很水啊……
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int n, m, s, a[100003], fa[100003], ans;
int tot, head[300003], nxt[300003], ver[300003];
inline void add(int u, int v)
{
ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}
void dfs(int u, int fa, int w)
{
if (w > s) return;
if (w == s) {++ans; return;}
for (int i = head[u]; i; i = nxt[i])
{
int v = ver[i];
if (v == fa) continue;
dfs(v, u, w + a[v]);
}
}
int main()
{
//File("B");
n = gi(), s = gi();
for (int i = 1; i <= n; i+=1) a[i] = gi();
for (int i = 1; i < n; i+=1)
{
int u = gi(), v = gI();
add(u, v);
fa[v] = u;
}
for (int i = 1; i <= n; i+=1)
{
dfs(i, fa[i], a[i]);
}
printf("%d\n", ans);
return 0;
}
C - 松鼠的新家
题解
倍增\(LCA+\)树上差分一遍过。
注意要判断两个房间是不是存在祖先关系。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <vector>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
itn n, a[300003], dep[300003], w[300003], fa[300003][23], sum, ans, tot, head[1200003], nxt[1200003], ver[1200003];
int f[300003], logn;
inline void add(int u, int v)
{
ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}
void dfs(int u, int baba, int step)
{
dep[u] = step; fa[u][0] = baba;
for (int i = 1; i <= logn; i+=1)
{
fa[u][i] = fa[fa[u][i - 1]][i - 1];
}
for (int i = head[u]; i; i = nxt[i])
{
int v = ver[i];
if (v == baba) continue;
dfs(v, u, step + 1);
}
}
inline void getlca(int u, int v)
{
if (dep[u] < dep[v]) swap(u, v);
int x = u, y = v;
while (dep[u] > dep[v])
{
for (int i = logn; i >= 0; i-=1)
{
if (dep[fa[u][i]] > dep[v]) u = fa[u][i];
}
u = fa[u][0];
}
if (u == v)
{
++w[x], --w[fa[u][0]]; return;
}
for (itn i = logn; i >= 0; i-=1)
{
if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
}
u = fa[u][0];
++w[x], ++w[y], --w[u], --w[fa[u][0]];
}
void getans(int u)
{
f[u] = w[u];
for (int i = head[u]; i; i = nxt[i])
{
int v = ver[i];
if (v == fa[u][0]) continue;
getans(v);
f[u] = f[u] + f[v];
}
}
int main()
{
//File("C");
n = gi();
while ((1 << logn) < n) ++logn;
for (int i = 1; i <= n; i+=1)
{
a[i] = gi();
}
for (int i = 1; i < n; i+=1)
{
int u = gi(), v = gi();
add(u, v); add(v, u);
}
add(0, a[1]);
dfs(0, 0, 1);
for (int i = 1; i < n; i+=1)
{
getlca(a[i], a[i + 1]);
}
getans(a[1]);
for (int i = 1; i <= n; i+=1)
{
printf("%d\n", (i == a[1]) ? (f[i]) : (f[i] - 1));
}
return 0;
}
总结
为什么每次时间内\(0AC\)，时间结束之后再切题呢？
思维要更活跃一点啊。
