题意：

给你一个带?的字符串S，和一个字符串T，问把?替换后最多能匹配多少次T？可以重叠匹配。

题解：

这种肯定是要DP的。

怎么DP呢？

AC自动机上的DP问题很多，这个也可以用AC自动机。

dp[i][j]表示当前在S串的i位置，在AC自动机的j状态时能完整匹配T的次数。

 当i是问号时枚举26个字母转移，不是问号直接转移到这个字符。

滚动数组一下即可。

当然KMP也是可以的，因为只有一个串。不过需要手动建一下next数组，跟AC自动机一样。

代码：

AC自动机版本：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <bitset>
#include <map>
#include <vector>
#include <stack>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <ctime>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif

#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, int> pii;

const int MOD = 1e9 + 7;
const double PI = acos (-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 2e5 + 5;

char s[MAXN], t[MAXN];

const int SIGMA_SIZE = 130;
const int MAXNODE = 2e5 + 100;
const int MAXS = 150 + 10;
struct ACautomata {


    int ch[MAXNODE][SIGMA_SIZE];
    int f[MAXNODE];    // fail函数
    int val[MAXNODE];  // 每个字符串的结尾结点都有一个非0的val
    int last[MAXNODE]; // 输出链表的下一个结点
    //int cnt[MAXS];
    int sz;

    void init() {
        sz = 1;
        memset(ch[0], 0, sizeof(ch[0]));
        // memset(cnt, 0, sizeof(cnt));
    }

    // 字符c的编号
    inline int idx(char c) {
        return c - 'a';
    }

    // 插入字符串。v必须非0
    void insert(char *s, int v) {
        int u = 0, n = strlen(s);
        for(int i = 0; i < n; i++) {
            int c = idx(s[i]);
            if(!ch[u][c]) {
                memset(ch[sz], 0, sizeof(ch[sz]));
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        val[u] = v;
    }


    // 计算fail函数
    void getFail() {
        queue<int> q;
        f[0] = 0;
        // 初始化队列
        for(int c = 0; c < SIGMA_SIZE; c++) {
            int u = ch[0][c];
            if(u) {
                f[u] = 0;
                q.push(u);
                last[u] = 0;
            }
        }
        // 按BFS顺序计算fail
        while(!q.empty()) {
            int r = q.front();
            q.pop();
            for(int c = 0; c < SIGMA_SIZE; c++) {
                int u = ch[r][c];
                if(!u) {
                    ch[r][c]=ch[f[r]][c];
                    continue;
                }
                q.push(u);
                int v = f[r];
                while(v && !ch[v][c]) v = f[v];
                f[u] = ch[v][c];
                last[u] = val[f[u]] ? f[u] : last[f[u]];
            }
        }
    }

} ac;

int d[2][MAXN];

int main() {
#ifdef LOCAL
    freopen ("input.txt", "r", stdin);
#endif
    scanf("%s %s", s + 1, t + 1);
    ac.init();
    ac.insert(t + 1, 1);
    ac.getFail();
    int n = strlen(s + 1), m = strlen(t + 1);
    memset(d, -1, sizeof(d));
    d[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) d[i & 1][j] = -1;
        for (int j = 0; j <= m; j++) {
            if(d[i & 1 ^ 1][j] == -1) continue;
            if (s[i] == '?') {
                for (int k = 0; k < 26; k++)
                    d[i & 1][ac.ch[j][k]] = max(d[i & 1 ^ 1][j] + (ac.ch[j][k] == m) , d[i & 1][ac.ch[j][k]]);
            } else
            d[i & 1][ac.ch[j][s[i] - 'a']] = max(d[i & 1 ^ 1][j] + (ac.ch[j][s[i] - 'a'] == m) , d[i & 1][ac.ch[j][s[i] - 'a']]);
        }
    }
    int ans = 0;
    for (int i = 0; i <= m; i++) ans = max(ans, d[n & 1][i]);
    printf("%d\n", ans);
    return 0;
}

KMP版本:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <bitset>
#include <map>
#include <vector>
#include <stack>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <ctime>
#ifdef LOCAL
#define debug(x) cout<<#x<<" = "<<(x)<<endl;
#else
#define debug(x) 1;
#endif

#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define lowbit(x) x&-x
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, int> pii;

const int MOD = 1e9 + 7;
const double PI = acos (-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 2e5 + 5;

char s[MAXN], t[MAXN];
int f[MAXN];

void getFail(char *P, int m) {
    f[0] = f[1] = 0;
    for (int i = 2; i <= m; i++) {
        int j = f[i - 1];
        while (j && P[i] != P[j + 1])
            j = f[j];
        f[i] = (P[i] == P[j + 1]) ? j + 1: 0;
    }
}

int ch[MAXN][26];
int d[2][MAXN];

int main() {
#ifdef LOCAL
    freopen ("input.txt", "r", stdin);
#endif
    scanf("%s %s", s + 1, t + 1);
    int n = strlen(s + 1), m = strlen(t + 1);
    getFail(t, m);
    memset(d, -1, sizeof(d));
    d[0][0] = 0;
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j < 26; j++) {
            if(t[i + 1] == j + 'a') ch[i][j] = i + 1;
            else ch[i][j] = ch[f[i]][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) d[i & 1][j] = -1;
        for (int j = 0; j <= m; j++) {
            if (d[i & 1 ^ 1][j] == -1) continue;
            if (s[i] == '?') {
                for (int k = 0; k < 26; k++)
                    d[i & 1][ch[j][k]] = max(d[i & 1 ^ 1][j] + (ch[j][k] == m) , d[i & 1][ch[j][k]]);
            } else
            d[i & 1][ch[j][s[i] - 'a']] = max(d[i & 1 ^ 1][j] + (ch[j][s[i] - 'a'] == m) , d[i & 1][ch[j][s[i] - 'a']]);
        }
    }
    int ans = 0;
    for (int i = 0; i <= m; i++) ans = max(ans, d[n & 1][i]);
    printf("%d\n", ans);
    return 0;
}