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

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 //最小回文分解
 //根据题目需求求出分解成怎样的回文串
 //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
 char s[maxn], s1[maxn], s2[maxn];

 struct Palindrome_Automaton {
     int len[maxn], next[maxn][], fail[maxn], cnt[maxn];
     int num[maxn], S[maxn], sz, n, last;
     int diff[maxn];//表示相邻回文后缀的等差；
     int slk[maxn];//表示上一个等差数列的末项
     int fp[maxn];

     int newnode(int l) {
         for (int i = ; i < ; ++i)next[sz][i] = ;
         cnt[sz] = num[sz] = , len[sz] = l;
         return sz++;
     }

     void init() {
         sz = n = last = ;
         newnode();
         newnode(-);
         S[] = -;
         fail[] = ;
     }

     int get_fail(int x) {
         while (S[n - len[x] - ] != S[n])x = fail[x];
         return x;
     }

     void add(int c, int pos) {
         c -= 'a';
         S[++n] = c;
         int cur = get_fail(last);
         if (!next[cur][c]) {
             int now = newnode(len[cur] + );
             fail[now] = next[get_fail(fail[cur])][c];
             next[cur][c] = now;
             num[now] = num[fail[now]] + ;
             diff[now] = len[now] - len[fail[now]];
             slk[now] = (diff[now] == diff[fail[now]] ? slk[fail[now]] : fail[now]);
         }
         last = next[cur][c];
         cnt[last]++;
     }

     //dp[i]表示s[1...i] s[1...i]s[1...i]的最少反转次数。
     //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。
     void solve(int n, int *dp, int *pre) {
         for (int i = ; i <= n; i++) dp[i] = INF, pre[i] = ;
         init();
         dp[] = , fp[] = ;
         for (int i = ; i <= n; i++) {
             add(s[i], i);
             for (int j = last; j; j = slk[j]) {
                 fp[j] = i - len[slk[j]] - diff[j];
                 if (diff[j] == diff[fail[j]] && dp[fp[j]] > dp[fp[fail[j]]]) fp[j] = fp[fail[j]];
                 if (i %  ==  && dp[i] > dp[fp[j]] + ) {//分解成 长度为偶数的回文串
                     dp[i] = dp[fp[j]] + ;
                     pre[i] = fp[j];
                 }
             }
             if (i %  ==  && s[i] == s[i - ] && dp[i] >= dp[i - ]) {//长度是2的表示没有反转
                 dp[i] = dp[i - ];
                 pre[i] = i - ;
             }
         }
     }
 } pam;

 int dp[maxn], pre[maxn];

 int main() {
     // FIN;
     sffs(s1 + , s2 + );
     int n =  * strlen(s1 + ), len1 = , len2 = ;
     for (int i = ; i <=  * n; i++) {
         if (i & ) s[i] = s1[++len1];
         else s[i] = s2[++len2];
     }
     // fuck(s + 1);
     pam.solve(n, dp, pre);
     if (dp[n] > n) return  * printf("-1\n");
     else printf("%d\n", dp[n]);
     for (int i = n; i; i = pre[i])
         if (i - pre[i] > ) printf("%d %d\n", pre[i] /  + , i / );
     return ;
 }