今天是这几天考试中唯一一次发挥正常一点儿的...
T1:https://www.luogu.org/problem/T93119
这么小的数据，O（n^3）暴力就行
而我非常愚蠢加个二分，一样可以过
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<climits>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define B cout << "breakpoint" << endl;
#define clr(a) memset(a,0,sizeof(a));
inline int read()
{
int ans = 0,op = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-') op = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
(ans *= 10) += ch - '0';
ch = getchar();
}
return ans * op;
}
const int maxn = 305;
int a1[maxn],a2[maxn];
int n,k;
char s[maxn];
inline bool check(int d)
{
for(int i = 1;i + d - 1<= n;i++)
for(int j = 1;j + d - 1 <= n;j++)
{
int cnt = 0;
for(int t = 0;t < d;t++)
if(a1[i + t] != a2[j + t]) cnt++;
if(cnt <= k) return 1;
}
return 0;
}
int main()
{
freopen("master.in","r",stdin);
freopen("master.out","w",stdout);
n = read(),k = read();
scanf("%s",s); for(int i = 1;i <= n;i++) a1[i] = (int)s[i - 1] - 'a' + 1;
scanf("%s",s); for(int i = 1;i <= n;i++) a2[i] = (int)s[i - 1] - 'a' + 1;
int l = 0,r = n;
while(l < r)
{
int mid = (l + r + 1) >> 1;
if(check(mid)) l = mid;
else r = mid - 1;
}
printf("%d",l);
}
/*
7 2
aabbaba
bababbc
*/
T2:https://www.luogu.org/problem/T93120
考场上自信以为自己写的正解，没想到复杂度算错...
下次以为切题的时候，一定要好好算复杂度
70pts：f[u][s]:位于点u，还剩s条边可走
枚举起点s，容易处理环的情况
O(N^3)
正解：不考虑环，对于一条边u->v,它对答案的贡献是(d[u] - 1) *(d[v] - 1)；
考虑去掉三元环，令s1为u能到达的点集，s2为v能到达的点集，这样对于s1与s2的并集，他们都可以成为三元环
于是统计并集中的点数，减去答案即可
用bitset记录点集，O(n^3 / 32)
bitset怎么用么...：https://www.cnblogs.com/zwfymqz/archive/2018/04/02/8696631.html
T3:https://www.luogu.org/problem/T93121
考场上Trie树乱搞，至今不知道为什么是错的...
正解是优化建边
对于每一个i，把val[i]拆出来,i->val[i]连一条边权为1的边，然后val[i]向val[i]去掉一个1的权点连边，边权为0
先bfs，对于队首u，先走u的边权为1的边，再dfs边权为0的边
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<climits>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define B cout << "breakpoint" << endl;
#define clr(a) memset(a,0,sizeof(a));
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
inline int read()
{
int ans = 0,op = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-') op = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
(ans *= 10) += ch - '0';
ch = getchar();
}
return ans * op;
}
const int maxn = 2e6 + 5;
struct edge
{
int to,next,cost;
}e[maxn << 1];
int fir1[maxn],alloc,fir2[maxn];
void adde(int u,int v,int op)
{
if(op == 1) e[++alloc].next = fir1[u], fir1[u] = alloc, e[alloc].to = v;
else e[++alloc].next = fir2[u],fir2[u] = alloc,e[alloc].to = v;
}
const int M = 1 << 20;
int dis[maxn];
queue<int> q;
int n,m;
void dfs(int u,int d)
{
if(dis[u] >= 0) return;
dis[u] = d;
q.push(u);
for(int i = fir2[u];i;i = e[i].next) dfs(e[i].to,d);
if(u > M) return;
for(int i = 0;i < 20;i++) if(u >> i & 1) dfs(u ^ (1 << i),d);
}
void bfs()
{
memset(dis,-1,sizeof(dis));
dis[M + 1] = 0;
q.push(M + 1);
dfs(M + 1,0);
while(q.size())
{
int u = q.front(); q.pop();
for(int i = fir1[u];i;i = e[i].next) dfs(e[i].to,dis[u] + 1);
}
for(int i = 1;i <= n;i++)
printf("%d\n",dis[M + i]);
}
int main()
{
freopen("walk.in","r",stdin);
freopen("walk.out","w",stdout);
n = read(),m = read();
for(int i = 1;i <= n;i++)
{
int x = read();
adde(M + i,x,1);
adde(x,M + i,2);
}
for(int i = 1;i <= m;i++)
{
int u = read(),v = read();
adde(u + M,v + M,1);
}
bfs();
}