「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)

# 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
;
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;
}

View Code
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。。