BZOJ 4032: [HEOI2015]最短不公共子串 (dp*3 + SAM)

转博客大法好

第4个子任务中,为什么只转移最近的一个位置,自己YY吧(多YY有益身体健康).

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
template<class T>inline void read(T &num) {
    register char ch; register int flg = 1;
    while(!isdigit(ch=getchar()))if(ch=='-')flg=-flg;
    for(num=0; isdigit(ch); num=num*10+ch-'0', ch=getchar());
    num *= flg;
}
const int MAXN = 2005;
const int C = 26;
char A[MAXN], B[MAXN];
int f[MAXN][MAXN];
namespace task1 {
    inline void solve() {
        int ans = MAXN-1;
        for(int i = 1; A[i]; ++i) {//字串的右端点
            int mx = 0;
            for(int j = 1; B[j]; ++j)
                if(A[i] == B[j])
                    mx = max(mx, f[i][j] = f[i-1][j-1] + 1);
            if(mx < i) ans = min(ans, mx+1);
        }
        printf("%d\n", (A[ans] && B[ans]) ? ans : -1);
    }
}
namespace task2 {
    inline void solve() {
        int ans = MAXN-1;
        for(int i = 1; A[i]; ++i) {//字串的右端点
            int mx = 0;
            for(int j = 1; B[j]; ++j) {
                if(A[i] == B[j])
                    mx = max(mx, f[i][j] = f[i-1][j-1] + 1);
                else mx = max(mx, f[i][j] = f[i][j-1]);
            }
            if(mx < i) ans = min(ans, mx+1);
        }
        printf("%d\n", (A[ans] && B[ans]) ? ans : -1);
    }
}
namespace task3 {
    int sz, last, len[MAXN<<1], link[MAXN<<1], ch[MAXN<<1][C], dp[MAXN<<1];
    inline void init() {
        memset(dp, 0x3f, sizeof dp);
        sz = last = 0; ++sz;
        len[0] = dp[0] = 0;
        link[0] = -1;
        //memset(ch, 0, sizeof ch);
    }
    inline void insert(int c) {
        int cur = sz++, p;
        len[cur] = len[last] + 1;
        for(p = last; ~p && !ch[p][c]; p = link[p]) ch[p][c] = cur;
        if(p == -1) link[cur] = 0;
        else {
            int q = ch[p][c];
            if(len[p] + 1 == len[q]) link[cur] = q;
            else {
                int clone = sz++;
                len[clone] = len[p] + 1;
                link[clone] = link[q];
                memcpy(ch[clone], ch[q], sizeof ch[q]);
                for(; ~p && ch[p][c] == q; p = link[p]) ch[p][c] = clone;
                link[cur] = link[q] = clone;
            }
        }
        last = cur;
    }
    inline void solve() {
        init();
        for(int i = 1; B[i]; ++i) insert(B[i]-'a');
        int ans = MAXN-1;
        for(int i = 1, c; A[i]; ++i) {
            c = A[i]-'a';
            for(int j = 0; j < sz; ++j) {
                if(ch[j][c]) dp[ch[j][c]] = min(dp[ch[j][c]], dp[j] + 1);
                else ans = min(ans, dp[j] + 1);
            }
        }
        printf("%d\n", (A[ans] && B[ans]) ? ans : -1);
    }
}
namespace task4 {
    int nxt[MAXN][C], pos[C], dp[MAXN];
    inline void solve() {
        int len = strlen(B+1);
        memset(dp, 0x3f, sizeof dp);
        dp[0] = 0;
        for(int i = len; ~i; --i) { //要循环到0
            for(int c = 0; c < 26; ++c)
                nxt[i][c] = pos[c];
            if(i) pos[B[i]-'a'] = i;
        }
        int ans = MAXN-1;
        for(int i = 1, c; A[i]; ++i) {
            c = A[i]-'a';
            for(int j = len; ~j; --j) //要循环到0
                if(nxt[j][c]) dp[nxt[j][c]] = min(dp[nxt[j][c]], dp[j] + 1);
                else ans = min(ans, dp[j] + 1);
        }
        printf("%d\n", (A[ans] && B[ans]) ? ans : -1);
    }
}
int main() {
    scanf("%s%s", A+1, B+1);
    task1::solve();
    task2::solve();
    task3::solve();
    task4::solve();
}