单独把每个字母第一次出现和最后一次出现拿出来处理一下就好
#include <cstdio> #include <cstring> #include <iostream> #include <map> #include <set> #include <vector> #include <string> #include <queue> #include <deque> #include <bitset> #include <list> #include <cstdlib> #include <climits> #include <cmath> #include <ctime> #include <algorithm> #include <stack> #include <sstream> #include <numeric> #include <fstream> #include <functional> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int,int> pii; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 5005; int n,m,f[maxn][maxn]; char str1[maxn], str2[maxn]; int scnt1[maxn][26], scnt2[maxn][26]; int main() { int T; scanf("%d",&T); while(T--) { scanf("%s%s",str1 + 1,str2 + 1); int len1 = strlen(str1 + 1), len2 = strlen(str2 + 1); memset(scnt1,0,sizeof(scnt1)); memset(scnt2,0,sizeof(scnt2)); for(int i = 1;i <= len1;i++) { int nc = str1[i] - 'A'; for(int j = 0;j < 26;j++) { if(j == nc) scnt1[i][j] = scnt1[i - 1][j] + 1; else scnt1[i][j] = scnt1[i - 1][j]; } } for(int i = 1;i <= len2;i++) { int nc = str2[i] - 'A'; for(int j = 0;j < 26;j++) { if(j == nc) scnt2[i][j] = scnt2[i - 1][j] + 1; else scnt2[i][j] = scnt2[i - 1][j]; } } for(int i = 0;i <= len1;i++) { for(int j = 0;j <= len2;j++) { f[i][j] = INF; } } f[0][0] = 0; for(int i = 0;i <= len1;i++) { for(int j = 0;j <= len2;j++) { int npos, nc, ccnt, zcnt; if(i < len1) { npos = i + 1 + j, nc = str1[i + 1] - 'A', ccnt = scnt1[i + 1][nc] + scnt2[j][nc], zcnt = scnt1[len1][nc] + scnt2[len2][nc]; if(ccnt == 1) { if(zcnt == 1) f[i + 1][j] = min(f[i + 1][j], f[i][j]); else { f[i + 1][j] = min(f[i + 1][j], f[i][j] - npos); } } else if(ccnt == zcnt) { f[i + 1][j] = min(f[i + 1][j], f[i][j] + npos); } else f[i + 1][j] = min(f[i + 1][j], f[i][j]); } if(j < len2) { npos = i + 1 + j; nc = str2[j + 1] - 'A'; ccnt = scnt1[i][nc] + scnt2[j + 1][nc]; zcnt = scnt1[len1][nc] + scnt2[len2][nc]; if(ccnt == 1) { if(zcnt == 1) f[i][j + 1] = min(f[i][j + 1], f[i][j]); else { f[i][j + 1] = min(f[i][j + 1], f[i][j] - npos); } } else if(ccnt == zcnt) { f[i][j + 1] = min(f[i][j + 1], f[i][j] + npos); } else f[i][j + 1] = min(f[i][j + 1], f[i][j]); } } } printf("%d ",f[len1][len2]); } return 0; }