「6月雅礼集训 2017 Day1」说无可说
【题目大意】
给出n个字符串,求有多少组字符串之间编辑距离为1~8。
n<=200,∑|S| <= 10^6
【题解】
首先找编辑距离有一个n^2的dp,由于发现只找小于等于8的,所以搜旁边16个状态即可。
复杂度O(n^2|S| * 16)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# include <vector> # include <stdio.h> # include <iostream> # include <string.h> # include <algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef long double ld; const int N = 200 + 10, M = 1e6 + 10, AN = 10; const int mod = 1e9 + 7; int n, len[M], ans[AN]; vector<char> a[N]; char str[M]; int u, v; int dp[2][M]; inline int f(int i, int j) { if(i == 0) return j; if(j == 0) return i; if(j < max(1, i-8) || j > min(len[v], i+8)) return 1e9; return dp[i&1][j]; } inline void fin(int i, int j, int s) { dp[i&1][j] = s; } inline int gans() { for (int i=1; i<=len[v] || i<=len[u]; ++i) fin(0, i, 0), fin(1, i, 0); for (int i=1; i<=len[u]; ++i) for (int j=max(1, i-8), jto = min(i+8, len[v]); j<=jto; ++j) { fin(i, j, min(f(i-1, j), f(i, j-1)) + 1); if(a[u][i-1] != a[v][j-1]) fin(i, j, min(f(i, j), f(i-1, j-1) + 1)); else fin(i, j, min(f(i, j), f(i-1, j-1))); } return f(len[u], len[v]); } int main() { // freopen("say.in", "r", stdin); // freopen("say.out", "w", stdout); cin >> n; for (int i=1; i<=n; ++i) { scanf("%s", str); len[i] = strlen(str); for (int j=0; str[j]; ++j) a[i].push_back(str[j]); } for (int i=1, tem; i<=n; ++i) { for (int j=i+1; j<=n; ++j) { u = i, v = j; tem = gans(); if(tem >= 1 && tem <= 8) ++ ans[tem]; } } for (int i=1; i<=8; ++i) cout << ans[i] << ' '; cout << endl; return 0; }
当然,过不了,有60分。
所以考虑dp有很多空余状态,改成dfs加剪枝就可以过了。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# include <vector> # include <stdio.h> # include <iostream> # include <string.h> # include <algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef long double ld; const int N = 200 + 10, M = 1e6 + 10, AN = 10; const int mod = 1e9 + 7; int n, len[M], ans[AN]; vector<char> a[N]; char str[M]; # define ABS(x) ((x) > 0 ? (x) : -(x)) int tem; inline void solve(int u, int v, int cur_u, int cur_v, int cnt) { if(cnt + ABS(len[v] - len[u] - (cur_v - cur_u)) >= tem) return; while(cur_u < len[u] && cur_v < len[v]) { if(a[u][cur_u] != a[v][cur_v]) { solve(u, v, cur_u+1, cur_v, cnt+1); solve(u, v, cur_u, cur_v+1, cnt+1); solve(u, v, cur_u+1, cur_v+1, cnt+1); return ; } ++cur_u, ++cur_v; } if(cur_u == len[u]) tem = min(tem, cnt + len[v] - cur_v); if(cur_v == len[v]) tem = min(tem, cnt + len[u] - cur_u); } int main() { // freopen("say.in", "r", stdin); // freopen("say.out", "w", stdout); cin >> n; for (int i=1; i<=n; ++i) { scanf("%s", str); len[i] = strlen(str); for (int j=0; str[j]; ++j) a[i].push_back(str[j]); } for (int i=1; i<=n; ++i) { for (int j=i+1; j<=n; ++j) { tem = 9; solve(i, j, 0, 0, 0); ++ ans[tem]; } } for (int i=1; i<=8; ++i) cout << ans[i] << ' '; cout << endl; return 0; }
这里dfs的时候需要注意一点,要一段一段跳,不能一格一格跳,会T。。