Reverses CodeForces - 906E （最小回文分解）

题意：

给你两个串s和t，其中t是由s中选择若干个不相交的区间翻转得到的，现在要求求出最少的翻转次数以及给出方案。 
1≤|s|=|t|≤500000

题解：

我们将两个字符串合成成T=s1t1s2t2...sntn T=s1t1s2t2...sntn

那么问题就是最少要把整个字符串T 拆分成若干个偶数长度（并且长度大于2）的回文串。
长度是2的表示没有反转。
然后就变成了最小回文分解模型 ，然后直接上板子。

最小回文分解 论文在此

  1 #include <set>
  2 #include <map>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <unordered_map>
 14
 15 #define  pi    acos(-1.0)
 16 #define  eps   1e-9
 17 #define  fi    first
 18 #define  se    second
 19 #define  rtl   rt<<1
 20 #define  rtr   rt<<1|1
 21 #define  bug                printf("******\n")
 22 #define  mem(a, b)          memset(a,b,sizeof(a))
 23 #define  name2str(x)        #x
 24 #define  fuck(x)            cout<<#x" = "<<x<<endl
 25 #define  sfi(a)             scanf("%d", &a)
 26 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 27 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 28 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 29 #define  sfL(a)             scanf("%lld", &a)
 30 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 31 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 32 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 33 #define  sfs(a)             scanf("%s", a)
 34 #define  sffs(a, b)         scanf("%s %s", a, b)
 35 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 36 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 37 #define  FIN                freopen("../in.txt","r",stdin)
 38 #define  gcd(a, b)          __gcd(a,b)
 39 #define  lowbit(x)          x&-x
 40 #define  IO                 iOS::sync_with_stdio(false)
 41
 42
 43 using namespace std;
 44 typedef long long LL;
 45 typedef unsigned long long ULL;
 46 const ULL seed = 13331;
 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 48 const int maxn = 1e6 + 7;
 49 const int maxm = 8e6 + 10;
 50 const int INF = 0x3f3f3f3f;
 51 const int mod = 1e9 + 7;
 52 //最小回文分解
 53 //根据题目需求求出分解成怎样的回文串
 54 //本代码用于将串分解成最小数目的长度偶数回文的方案数
 55 char s[maxn], s1[maxn], s2[maxn];
 56
 57 struct Palindrome_Automaton {
 58     int len[maxn], next[maxn][26], fail[maxn], cnt[maxn];
 59     int num[maxn], S[maxn], sz, n, last;
 60     int diff[maxn];//表示相邻回文后缀的等差；
 61     int slk[maxn];//表示上一个等差数列的末项
 62     int fp[maxn];
 63
 64     int newnode(int l) {
 65         for (int i = 0; i < 26; ++i)next[sz][i] = 0;
 66         cnt[sz] = num[sz] = 0, len[sz] = l;
 67         return sz++;
 68     }
 69
 70     void init() {
 71         sz = n = last = 0;
 72         newnode(0);
 73         newnode(-1);
 74         S[0] = -1;
 75         fail[0] = 1;
 76     }
 77
 78     int get_fail(int x) {
 79         while (S[n - len[x] - 1] != S[n])x = fail[x];
 80         return x;
 81     }
 82
 83     void add(int c, int pos) {
 84         c -= ‘a‘;
 85         S[++n] = c;
 86         int cur = get_fail(last);
 87         if (!next[cur][c]) {
 88             int now = newnode(len[cur] + 2);
 89             fail[now] = next[get_fail(fail[cur])][c];
 90             next[cur][c] = now;
 91             num[now] = num[fail[now]] + 1;
 92             diff[now] = len[now] - len[fail[now]];
 93             slk[now] = (diff[now] == diff[fail[now]] ? slk[fail[now]] : fail[now]);
 94         }
 95         last = next[cur][c];
 96         cnt[last]++;
 97     }
 98
 99     //dp[i]表示s[1...i] s[1...i]s[1...i]的最少反转次数。
100     //pre[i] pre[i]pre[i]表示在最优解里面i为区间右端点的左端点下标。
101     void solve(int n, int *dp, int *pre) {
102         for (int i = 0; i <= n; i++) dp[i] = INF, pre[i] = 0;
103         init();
104         dp[0] = 0, fp[0] = 1;
105         for (int i = 1; i <= n; i++) {
106             add(s[i], i);
107             for (int j = last; j; j = slk[j]) {
108                 fp[j] = i - len[slk[j]] - diff[j];
109                 if (diff[j] == diff[fail[j]] && dp[fp[j]] > dp[fp[fail[j]]]) fp[j] = fp[fail[j]];
110                 if (i % 2 == 0 && dp[i] > dp[fp[j]] + 1) {//分解成 长度为偶数的回文串
111                     dp[i] = dp[fp[j]] + 1;
112                     pre[i] = fp[j];
113                 }
114             }
115             if (i % 2 == 0 && s[i] == s[i - 1] && dp[i] >= dp[i - 2]) {//长度是2的表示没有反转
116                 dp[i] = dp[i - 2];
117                 pre[i] = i - 2;
118             }
119         }
120     }
121 } pam;
122
123 int dp[maxn], pre[maxn];
124
125 int main() {
126     // FIN;
127     sffs(s1 + 1, s2 + 1);
128     int n = 2 * strlen(s1 + 1), len1 = 0, len2 = 0;
129     for (int i = 1; i <= 2 * n; i++) {
130         if (i & 1) s[i] = s1[++len1];
131         else s[i] = s2[++len2];
132     }
133     // fuck(s + 1);
134     pam.solve(n, dp, pre);
135     if (dp[n] > n) return 0 * printf("-1\n");
136     else printf("%d\n", dp[n]);
137     for (int i = n; i; i = pre[i])
138         if (i - pre[i] > 2) printf("%d %d\n", pre[i] / 2 + 1, i / 2);
139     return 0;
140 }

原文地址：https://www.cnblogs.com/qldabiaoge/p/11403882.html