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Sol

神仙題Orz

考慮區間dp，如果我們只設\(f[l][r]\)表示\(s_{lr}\)被壓縮的最小長度，而不去關心內部\(M\)分布的話，可能在轉移的時候轉移出非法狀態

因此考慮多加一維表示當前子串中有沒有\(M\)(默認第一個字符為\(M\)不統計在內)

轉移的時候就考慮不同的\(M\)對當前區間的貢獻就可以。

\(P\)的作用實際上是將兩個相同的字符串合成一個，拿hash判一下

復雜度\(O(n^3)\)

`cpp #include<bits/stdc++.h> #define Pair pair<int, int> #define MP(x, y) make_pair(x, y) #define fi first #define se second #define LL long long #define ull unsigned long long #define Fin(x) {freopen(#x".in","r",stdin);} #define Fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int MAXN = 501, mod = 1e9 + 7, INF = 1e9 + 10; const double eps = 1e-9; template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;} template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;} template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;} template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;} template <typename A> inline void debug(A a){cout << a << ‘\n‘;} template <typename A> inline LL sqr(A x){return 1ll * x * x;} inline int read() { char c = getchar(); int x = 0, f = 1; 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; } int N, f[MAXN][MAXN], num[MAXN]; char s[MAXN]; ull po[MAXN], ha[MAXN], base = 131; ull get(int l, int r) { return ha[r] - ha[l - 1] * po[r - l + 1]; } signed main() { scanf("%s", s + 1); N = strlen(s + 1); memset(f, 0x3f, sizeof(f)); po[0] = 1; for(int i = 1; i <= N; i++) f[i][i] = 1, num[i] = num[i / 10] + 1, ha[i] = ha[i - 1] * base + s[i], po[i] = po[i - 1] * base; for(int len = 2; len <= N; len++) { for(int l = 1; l + len - 1 <= N; l++) { int r = l + len - 1; for(int cur = 1; cur <= len; cur++) { if(len % cur == 0) { bool flag = 1; for(int i = l; i + 2 * cur - 1 <= r; i++) if(get(i, i + cur - 1) != get(i + cur, i + 2 * cur - 1)) {flag = 0; break;} if(flag) chmin(f[l][r], f[l][l + cur - 1] + num[len / cur] + 2); } } for(int k = l; k < r; k++) chmin(f[l][r], f[l][k] + f[k + 1][r]); } } cout << f[1][N]; return 0; } /* 20 1 8 4 4 6 7 4 4 0 7 3 7 0 9 5 5 1 6 1 8 */ }

洛谷P2470 [SCOI2007]壓縮(區間dp)