cf1043D. Mysterious Crime(枚举)

题意

给出\(m\)个长度为\(n\)的排列，问有多少连续公共子串

\(m \leqslant 10, n \leqslant 10^5\)

Sol

非常naive的一道题然而交了3遍才过(昨晚真的困得不行。。)

枚举第一个串的位置，直接维护10个指针，分别表示每个串与第一个串匹配到的位置，同时记录出已经匹配了多少位

因为题目中给出的是排列，所以如果某一位不能匹配了，则需要从新的位置开始

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<map>

#include<vector>

#include<set>

#include<queue>

#include<cmath>

//#include<ext/pb_ds/assoc_container.hpp>

//#include<ext/pb_ds/hash_policy.hpp>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

//#define int long long

#define LL long long

#define ull unsigned long long

#define rg register

#define pt(x) printf("%d ", x);

//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)

//char buf[(1 << 22)], *p1 = buf, *p2 = buf;

//char obuf[1<<24], *O = obuf;

//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}

//#define OS  *O++ = ' ';

using namespace std;

//using namespace __gnu_pbds;

const int MAXN = 1e5 + 10, INF = 1e9 + 10, mod = 998244353;

const double eps = 1e-9;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}

	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

	return x * f;

}

int N, M;

int a[11][MAXN], pos[11][MAXN], now[MAXN], len[MAXN];

main() {

	N = read(); M = read();

	for(int i = 1; i <= M; i++)

		for(int j = 1; j <= N; j++) {

			a[i][j] = read();

			pos[i][a[i][j]] = j;

		}

	LL ans = 0;

	for(int i = 1; i <= N; i++) {

		len[1] = i;

		for(int j = 2; j <= M; j++) {

			if(a[j][now[j] + 1] != a[1][i]) now[j] = pos[j][a[1][i]], len[j] = 1;

			else now[j]++, len[j]++;

		}

		LL mn = INF;

		for(int j = 1; j <= M; j++) mn = min(mn, (LL)len[j]);

		ans += mn;

	}

	cout << ans;

	return 0;

}

/*

*/

巴特西

cf1043D. Mysterious Crime(枚举)

题意

Sol

最新文章

热门文章