Omkar has a pie tray with ? (2≤?≤20) spots. Each spot in the tray contains either a chocolate pie or a pumpkin pie. However, Omkar does not like the way that the pies are currently arranged, and has another ideal arrangement that he would prefer instead.
To assist Omkar, ? elves have gathered in a line to swap the pies in Omkar’s tray. The ?-th elf from the left is able to swap the pies at positions ?? and ?? in the tray.
In order to get as close to his ideal arrangement as possible, Omkar may choose a contiguous subsegment of the elves and then pass his pie tray through the subsegment starting from the left. However, since the elves have gone to so much effort to gather in a line, they request that Omkar’s chosen segment contain at least ? (1≤?≤?) elves.
Formally, Omkar may choose two integers ? and ? satisfying 1≤?≤?≤? and ???+1≥? so that first the pies in positions ?? and ?? will be swapped, then the pies in positions ??+1 and ??+1 will be swapped, etc. until finally the pies in positions ?? and ?? are swapped.
Help Omkar choose a segment of elves such that the amount of positions in Omkar’s final arrangement that contain the same type of pie as in his ideal arrangement is the maximum possible. Note that since Omkar has a big imagination, it might be that the amounts of each type of pie in his original arrangement and in his ideal arrangement do not match.
Input
The first line contains three integers ?, ?, and ? (1≤?≤?≤106 and 2≤?≤20) — the number of elves, the minimum subsegment length, and the number of spots in Omkar’s tray respectively.
The second and third lines each contain a string of length ? consisting of 0s and 1s that represent initial arrangement of pies and ideal arrangement of pies; the ?-th character in each string is equal to 0 if the ?-th spot in the arrangement contains a chocolate pie and is equal to 1 if the ?-th spot in the arrangement contains a pumpkin pie. It is not guaranteed that the two strings have the same amount of 0s or the same amount of 1s.
? lines follow. The ?-th of these lines contains two integers ?? and ?? (1≤??,??≤?, ??≠??) which indicate that the ?-th elf from the left can swap the pies at positions ?? and ?? in the tray.
Output
Output two lines.
The first line should contain a single integer ? (0≤?≤?) equal to the amount of positions that contain the same type of pie in Omkar’s final arrangement and in Omkar’s ideal arrangement; ? should be the maximum possible.
The second line should contain two integers ? and ? satisfying 1≤?≤?≤? and ???+1≥?, indicating that Omkar should pass his tray through the subsegment ?,?+1,…,? to achieve a final arrangement with ? positions having the same type of pie as his ideal arrangement.
If there are multiple answers you may output any of them.
Examples
inputCopy
4 2 5
11000
00011
1 3
3 5
4 2
3 4
outputCopy
5
1 3
inputCopy
4 3 5
11000
00011
1 3
1 5
2 4
1 5
outputCopy
3
1 4
Note
In the first test case, the swaps will go like this:
Swap 1 and 3: 11000 becomes 01100
Swap 3 and 5: 01100 becomes 01001
Swap 4 and 2: 01001 becomes 00011
The final arrangement is the same as the ideal arrangement 00011, so there are 5 positions with the same type of pie, which is optimal.
In the second test case, the swaps will go like this:
Swap 1 and 3: 11000 becomes 01100
Swap 1 and 5: 01100 becomes 01100
Swap 4 and 2: 01100 becomes 00110
Swap 1 and 5: 00110 becomes 00110
The final arrangement has 3 positions with the same type of pie as the ideal arrangement 00011, those being positions 1, 2, and 4. In this case the subsegment of elves (?,?)=(2,3) is more optimal, but that subsegment is only length 2 and therefore does not satisfy the constraint that the subsegment be of length at least ?=3.
参考题解
题意：
两个序列s,t。
然后一堆对s序列的交换操作。要求取连续且至少为m个的交换操作，使得s序列和t序列相同位置相同值个数最多。
思路：
通过本题学习了一下SOS DP，实际就是高效算二进制子集前缀和的算法。
本题需要找一段(l,r)(l,r)(l,r)交换操作，而我们只关心交换完了以后sss序列和ttt序列的相似度。所以可以假设是将sss序列按照(1,l?1)(1,l-1)(1,l?1)的操作进行交换，将ttt按照(1,r)(1,r)(1,r)的部分进行交换。这时候再算sss和ttt的相似度，就等价于sss按照(l,r)(l,r)(l,r)交换和ttt。
那么可以定义通过(1,x)(1,x)(1,x)操作dp[0][i]dp[0][i]dp[0][i]为将sss序列交换成s′s's′，且i∈s′i∈s'i∈s′的最小xxx。
通过(1,y)(1,y)(1,y)操作dp[1][i]dp[1][i]dp[1][i]为将ttt序列交换成t′t't′，且i∈t′i∈t'i∈t′的最大yyy。
那么只要满足dp[1][i]?dp[0][i]≥mdp[1][i]-dp[0][i]≥mdp[1][i]?dp[0][i]≥m，就说明这个序列可以被选择，且最后sss和ttt的相似部分就是iii，只要算出iii中的1，就可以知道相似度大小，为k?(o1+o2?num(i))k-(o1+o2-num(i))k?(o1+o2?num(i))，o1,o2o1,o2o1,o2为s,ts,ts,t中1的个数。
#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;typedef long long ll;const int maxn = 21;
const int INF = 0x3f3f3f3f;int n,m,k;int p[maxn],dp[2][1 << maxn];
char s[maxn],t[maxn];void get(int &x,int &y) {
x = y = 0;for(int i = 0;i < k;i++) {
if(s[i] == '1') x |= 1 << p[i];if(t[i] == '1') y |= 1 << p[i];}
}int num(int x) {
int cnt = 0;while(x) {
cnt++;x &= (x - 1);}return cnt;
}int main() {
scanf("%d%d%d",&n,&m,&k);scanf("%s%s",s,t);int o1 = 0,o2 = 0;for(int i = 0;i < k;i++) {
p[i] = i;o1 += (s[i] == '1');o2 += (t[i] == '1');}memset(dp[0],INF,sizeof(dp[0]));memset(dp[1],0,sizeof(dp[1]));int x = 0,y = 0;get(x,y);dp[0][x] = 0;dp[1][y] = 0;for(int i = 1;i <= n;i++) {
int a,b;scanf("%d%d",&a,&b);a--;b--;swap(p[a],p[b]);get(x,y);dp[0][x] = min(dp[0][x],i);dp[1][y] = max(dp[1][y],i);}int ans = 0,ansl = 1,ansr = 1;for(int i = (1 << k) - 1;i >= 0;i--) {
int tmp = k - (o1 + o2 - 2 * num(i));if(dp[1][i] - dp[0][i] >= m) {
if(ans < tmp) {
ans = tmp;ansl = dp[0][i] + 1;ansr = dp[1][i];}}for(int j = 0;j < k;j++) {
if((i >> j) & 1) {
dp[0][i ^ (1 << j)] = min(dp[0][i ^ (1 << j)],dp[0][i]);dp[1][i ^ (1 << j)] = max(dp[1][i ^ (1 << j)],dp[1][i]);}}}printf("%d\n",ans);printf("%d %d\n",ansl,ansr);return 0;
}