日常训练2025-1-12
P2679 [NOIP2015 提高组] 子串
普及+/提高
https://www.luogu.com.cn/problem/P2679
思路
https://www.luogu.com.cn/article/k0zkdin9
评述
做DP时可以把能想到的有用的状态都定义出来，后序在把不需要的，或者可以根据其他状态推出来的状态删除，逐渐优化。
代码
#include <bits/stdc++.h>typedef std::pair<long long, long long> pll;
typedef std::pair<int, int> pii;
#define INF 0x3f3f3f3f
#define MOD 1000000007
using i64 = long long;
const int N = 1e5+5;void solve(){int n, m, k;std::cin >> n >> m >> k;std::string a, b;std::cin >> a >> b;a = ' ' + a;b = ' ' + b;std::vector dp(m+1, std::vector<std::vector<int>>(k+1, std::vector<int>(2)));std::vector ndp(m+1, std::vector<std::vector<int>>(k+1, std::vector<int>(2))); dp[0][0][0] = 1;ndp[0][0][0] = 1;for (int i = 1; i <= n; i++){ndp = dp;for (int j = 1; j <= m; j++){for (int p = 1; p <= k; p++){if (a[i] == b[j]){dp[j][p][1] = (ndp[j-1][p][1] + ndp[j-1][p-1][1]) % MOD + ndp[j-1][p-1][0] % MOD;dp[j][p][1] %= MOD;dp[j][p][0] = (ndp[j][p][1] + ndp[j][p][0]) % MOD;}else{dp[j][p][1] = 0;dp[j][p][0] = (ndp[j][p][1] + ndp[j][p][0]) % MOD;}}}}std::cout << (dp[m][k][0] + dp[m][k][1]) % MOD << '\n';
}signed main()
{std::ios::sync_with_stdio(false);std::cin.tie(nullptr);std::cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(2);int t = 1, i;for (i = 0; i < t; i++){solve();}return 0;
}
