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模拟赛的时候强行迭代 dp 做的题, 事实上是 AC 自动机的套路题, 虽说迭代不用算法比较亲民, 但是确实抠细节不太好写, 当时也写了有一个小时, 如果直到这是一道 AC 自动机的套路题的话, 就好做多了
将 N + 1 个模式串插入 AC 自动机之后, 如果进入到末尾结点就在贡献处加上 (1LL <<M - i + 1) * 方案数的贡献, 表示当前长度有多少种可能到达这个节点, 同时到达了之后后面的所有节点都可以任选.
- #include <map>
- #include <set>
- #include <ctime>
- #include <cmath>
- #include <queue>
- #include <stack>
- #include <vector>
- #include <string>
- #include <bitset>
- #include <cstdio>
- #include <cstdlib>
- #include <cstring>
- #include <sstream>
- #include <iostream>
- #include <algorithm>
- #include <functional>
- using namespace std;
- #define For(i, x, y) for(int i=x;i<=y;i++)
- #define _For(i, x, y) for(int i=x;i>=y;i--)
- #define Mem(f, x) memset(f,x,sizeof(f))
- #define Sca(x) scanf("%d", &x)
- #define Sca2(x,y) scanf("%d%d",&x,&y)
- #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
- #define Scl(x) scanf("%lld",&x)
- #define Pri(x) printf("%d\n", x)
- #define Prl(x) printf("%lld\n",x)
- #define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
- #define LL long long
- #define ULL unsigned long long
- #define mp make_pair
- #define PII pair<int,int>
- #define PIL pair<int,long long>
- #define PLL pair<long long,long long>
- #define pb push_back
- #define fi first
- #define se second
- typedef vector<int> VI;
- int read(){int x = 0,f = 1;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 x*f;}
- const double PI = acos(-1.0);
- const double eps = 1e-9;
- const int maxn = 2010;
- const int INF = 0x3f3f3f3f;
- const int mod = 1e9 + 7;
- int N,M,K;
- char str[maxn];
- int nxt[maxn][26],tot,fail[maxn],ed[maxn],root;
- int newnode(){
- for(int i = 0 ; i <2; i ++) nxt[tot][i] = -1;
- ed[tot++] = 0;
- return tot - 1;
- }
- void init(){
- tot = 0;
- root = newnode();
- }
- void insert(char *str){
- int p = root;
- for(int i = 0; str[i]; i ++){
- int id = str[i] - '0';
- if(nxt[p][id] == -1) nxt[p][id] = newnode();
- p = nxt[p][id];
- }
- ed[p] = 1;
- }
- void Build(){
- queue<int>Q;
- fail[root] = root;
- for(int i = 0;i < 2; i ++){
- if(~nxt[root][i]){
- fail[nxt[root][i]] = root;
- Q.push(nxt[root][i]);
- }else{
- nxt[root][i] = root;
- }
- }
- while(!Q.empty()){
- int u = Q.front(); Q.pop();
- for(int i = 0 ; i < 2; i ++){
- if(~nxt[u][i]){
- fail[nxt[u][i]] = nxt[fail[u]][i];
- ed[nxt[u][i]] |= ed[nxt[fail[u]][i]];
- Q.push(nxt[u][i]);
- }else{
- nxt[u][i] = nxt[fail[u]][i];
- }
- }
- }
- }
- LL dp[2][maxn];
- int main(){
- int T = read();
- while(T--){
- Sca2(N,M); init();
- scanf("%s",str); insert(str);
- for(int i = 0 ; i < N ; i ++){
- if(str[i] == '1') str[i] = '0';
- else str[i] = '1';
- insert(str);
- if(str[i] == '1') str[i] = '0';
- else str[i] = '1';
- }
- Build();
- LL ans = 0;
- dp[0][0] = 1;
- for(int i = 1 ; i <= M ; i ++){
- for(int j = 0 ; j < tot; j ++) dp[i & 1][j] = 0;
- for(int j = 0 ; j < tot; j ++){
- if(ed[j]) continue;
- for(int k = 0 ; k < 2; k ++){
- if(ed[nxt[j][k]]){
- ans += dp[i + 1 & 1][j] * (1LL << (M - i));
- }else{
- dp[i & 1][nxt[j][k]] += dp[i + 1 & 1][j];
- }
- }
- }
- }
- Prl(ans);
- for(int j = 0 ; j < tot; j ++) dp[0][j] = dp[1][j] = 0;
- }
- return 0;
- }
来源: http://www.bubuko.com/infodetail-3184877.html