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HDU 5903 Square Distance (贪心+DP)
题意：一个字符串被称为square当且仅当它可以由两个相同的串连接而成. 例如, "abab", "aa"是square, 而"aaa", "abba"不是. 两个长度相同字符串之间的
hamming distance是对应位置上字符不同的位数. 给定一行字符串和 m，输出字典序最小的字符串。
析：首先先用dp判断能不能形成这样的字符串，然后再打印出来，dp[i][j] 表示 i - 中间的数能不能改 j 个字符得到，最后打印时贪心，从最小的开始，如果能用结束。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}int dp[maxn/2][maxn];char s[maxn], ans[maxn];int main(){ int T; cin >> T; while(T--){ scanf("%d %d", &n, &m); scanf("%s", s+1); memset(dp, false, sizeof dp); int mid = n/2; dp[mid+1][0] = true; for(int i = mid; i >= 1; --i){ if(s[i] == s[n-mid+i]){ for(int j = 0; j <= m; ++j){ if(j > 1) dp[i][j] |= dp[i+1][j-2]; dp[i][j] |= dp[i+1][j]; } } else{ for(int j = 0; j <= m; ++j){ if(j) dp[i][j] |= dp[i+1][j-1]; if(j > 1) dp[i][j] |= dp[i+1][j-2]; } } } if(!dp[1][m]) puts("Impossible"); else{ int cnt = m; for(int i = 1; i <= mid; ++i){ for(int j = 0; j < 26; ++j){ if(s[i] == ‘a‘+j || s[i+mid] == ‘a‘+j){ if(s[i] == s[i+mid]){ if(dp[i+1][cnt]) { ans[i] = ans[i+mid] = s[i]; break; } } else{ if(dp[i+1][cnt-1]){ ans[i] = ans[i+mid] = ‘a‘ + j; --cnt; break; } } } else{ if(dp[i+1][cnt-2]){ ans[i] = ans[i+mid] = ‘a‘ + j; cnt -= 2; break; } } } } ans[n+1] = 0; puts(ans+1); } } return 0;}
HDU 5903 Square Distance (贪心+DP)
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