• 洛谷P5108 仰望半月的夜空(后缀数组)


    题意

    题目链接

    Sol

    warning:下面这个做法只有95分,本地拍了1w+组都没找到错误我表示十分无能为力

    我们考虑每个串的排名去更新答案,显然排名为(1)的后缀的前缀一定是当前长度的字典序最小的答案,但不一定是左端点最小的答案,因此还需要用一个数据结构去维护一下所有可行的左端点。然后枚举所有后缀更新答案就行了。

    #include<bits/stdc++.h> 
    using namespace std;
    const int MAXN = 2e6 + 10, SS = 6e5 + 10, INF = 1e9 + 10;
    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;}
    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 CharSet, N, M, ans[SS];
    int s[SS]; char stmp[SS];
    int rak[SS], tp[SS], tax[MAXN * 10], sa[SS], height[SS];
    void Sort() {
        for(int i = 0; i <= M; i++) tax[i] = 0;
        for(int i = 1; i <= N; i++) tax[rak[i]]++;
        for(int i = 1; i <= M; i++) tax[i] += tax[i - 1];
        //for(int i = N; i >= 1; i--) sa[i] = tp[tax[rak[i]]--];
        for(int i = N; i >= 1; i--) sa[tax[rak[tp[i]]]--] = tp[i];
    }
    void SuffixArraryBuild() {
        M = CharSet;
        for(int i = 1; i <= N; i++) rak[i] = s[i], tp[i] = i;
     	Sort();
     	for(int w = 1, p; w < N; w <<= 1, M = p) {
     		p = 0;
     		for(int i = N - w + 1; i <= N; i++) tp[++p] = i;
     		for(int i = 1; i <= N; i++) if(sa[i] > w) tp[++p] = sa[i] - w;
            Sort(); swap(rak, tp);
            rak[sa[1]] = p = 1; 
            for(int i = 2; i <= N; i++) 
                rak[sa[i]] = (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w] ? p : ++p);
        }
        for(int i = 1, k = 0; i <= N; i++) {
        
            int j = sa[rak[i] - 1];
            if(k) k--;
            while(j && s[i + k] == s[j + k]) k++;//mmp一开始没写j.. 
                height[rak[i]] = k;//tag
        }
    }
    #define ls k << 1
    #define rs k << 1 | 1
    int mn[MAXN];
    void update(int k) {
        mn[k] = min(mn[ls], mn[rs]);
    }
    void Build(int k, int l, int r) {
        if(l == r) {mn[k] = sa[l]; return ;}
        int mid = l + r >> 1;
        Build(ls, l, mid); Build(rs, mid + 1, r);
        update(k);
    }
    int IntQuery(int k, int l, int r, int ql, int qr) {
        if(ql <= l && r <= qr) return mn[k];
        int mid = l + r >> 1;
        if(ql > mid) return IntQuery(rs, mid + 1, r, ql, qr);
        else if(qr <= mid) return IntQuery(ls, l, mid, ql, qr);
        else return min(IntQuery(ls, l, mid, ql, qr), IntQuery(rs, mid + 1, r, ql, qr));
    }
    int f[MAXN][20];
    int Find(int pos, int cur) {//从pos开始的height,第一个 < cur的位置 
        if(height[pos] < cur) return pos - 1;
        for(int i = 19; i >= 1; i--) {
            while(f[pos][i] >= cur && f[pos][i] <= INF && pos + (1 << i) - 1 <= N) 
                pos = pos + (1 << i) - 1;
        }
        return pos;
    }
    
    void solve() {
        Build(1, 1, N);
        memset(f, 0x3f, sizeof(f));
        for(int i = 1; i <= N; i++) f[i][0] = height[i];
        for(int j = 1; j <= 19; j++) 
            for(int i = 1; i + (1 << j) - 1<= N; i++) 
                chmin(f[i][j], min(f[i][j - 1], f[i + (1 << j - 1)][j - 1]));
        
        
        int cur = 1;//马上要找的答案 
        for(int rk = 1; rk <= N; rk++) {
            if((N - sa[rk] + 1) < cur) continue;
            int pos = sa[rk];
            for(int i = pos + cur - 1; i <= N; i++) {
                int j = Find(rk + 1, cur);
                ans[cur++] = IntQuery(1, 1, N, rk, j);
            }
        }
    }
    signed main() {
        CharSet = read(); N = read();
        if(CharSet == 26) {
            scanf("%s", stmp + 1);
            for(int i = 1; i <= N; i++) s[i] = stmp[i] - 'a';
        }
        else for(int i = 1; i <= N; i++) s[i] = read();
        SuffixArraryBuild();
        solve();
        for(int i = 1; i <= N; i++) cout << ans[i] << ' ';
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10612560.html
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