CSP-S开小灶4
作者:互联网
A. 山洞
暴力枚举 \(nm\) 暴力
然后发现枚举的步长每 \(n\) 个一循环
搞出从\(0, 0\) 走 \(n\) 步的数组
进而发现从 \(0\) 到 \(i\) 和从 \(x\) 到 \((x + i) \% n\) 等价
于是可以\(n^2\) 转移出走 \(n + n\) 步,然后发现可以倍增了
其实这本质上是个循环矩阵
我太菜了,今天才知道啥是循环矩阵
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1005;
const int mod = 1e9 + 7;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9') c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
return x;
}
int f[2][maxn], b[21][maxn], ans[maxn], tmp[maxn];
int n, m;
int main(){
n = read(); m = read();
f[0][0] = 1;
int zt = 1;
for(int i = 1; i <= min(n, m); ++i, zt = 1 - zt)
for(int p = 0; p < n; ++p)
if(f[1 - zt][p]){
int p1 = (p + i) % n, p2 = ((p - i) % n + n) % n;
f[zt][p1] = (f[zt][p1] + f[1 - zt][p]) % mod;
if(p1 != p2)f[zt][p2] = (f[zt][p2] + f[1 - zt][p]) % mod;
f[1 - zt][p] = 0;
}
if(m <= n){
printf("%d\n",f[m & 1][0]);
return 0;
}
for(int i = 0; i < n; ++i)b[0][i] = f[n & 1][i];
int round = m / n; ans[0] = 1;
if(round & 1)for(int i = 0; i < n; ++i)ans[i] = b[0][i];
for(int j = 1; (1 << j) <= round; ++j){
for(int i = 0; i < n; ++i)if(b[j - 1][i]){
for(int k = 0; k < n; ++k){
b[j][(i + k) % n] = (b[j][(i + k) % n] + 1ll * b[j - 1][i] * b[j - 1][k] % mod) % mod;
}
}
if(round & (1 << j)){
for(int i = 0; i < n; ++i)
for(int k = 0; k < n; ++k){
tmp[(i + k) % n] = (tmp[(i + k) % n] + 1ll * ans[i] * b[j][k] % mod) % mod;
}
for(int i = 0; i < n; ++i)ans[i] = tmp[i];
for(int i = 0; i < n; ++i)tmp[i] = 0;
}
}
int res = m % n;
for(int i = 1; i <= res; ++i){
for(int p = 0; p < n; ++p)if(ans[p]){
int p1 = (p + i) % n, p2 = ((p - i) % n + n) % n;
tmp[p1] = (tmp[p1] + ans[p]) % mod;
if(p1 != p2)tmp[p2] = (tmp[p2] + ans[p]) % mod;
ans[p] = 0;
}
for(int p = 0; p < n; ++p)ans[p] = tmp[p];
for(int p = 0; p < n; ++p)tmp[p] = 0;
}
printf("%d\n",ans[0]);
return 0;
}
B. beauty
对每条边的贡献分开考虑
\(\sum_{i = 1}^{n - 1} min(cnt_i, k +k - cnt_i)\)
感觉最近想 \(DP\) 有点魔怔了
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100005;
inline int read(){
int x = 0; char c = getchar();
while(c < '0' || c > '9') c = getchar();
do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
return x;
}
int n, k;
int key[maxn];
vector<int>g[maxn];
int ans;
void dfs(int x, int fa){
for(int v : g[x]){
if(v == fa)continue;
dfs(v, x);
ans = (ans + min(key[v], k + k - key[v]));
key[x] += key[v];
}
}
int main(){
n = read(), k = read(); read();
for(int i = 1; i <= k + k; ++i)key[read()] = 1;
for(int i = 1; i < n; ++i){
int u = read(), v = read();
g[u].push_back(v); g[v].push_back(u);
}
dfs(1, 0);
printf("%d\n",ans);
return 0;
}
标签:int,long,read,maxn,key,ans,开小灶,CSP 来源: https://www.cnblogs.com/Chencgy/p/16698066.html