8.9 Round 1

Posted 2022-06-24 lm-lbg

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
*/

View Code

 

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();

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