• Mysterious Crime CodeForces


    大意: 给定m个n排列, 求有多少个公共子串.

    枚举每个位置, hash求出最大匹配长度. 

    #include <iostream>
    #include <sstream>
    #include <algorithm>
    #include <cstdio>
    #include <math.h>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #include <string.h>
    #include <bitset>
    #define REP(i,a,n) for(int i=a;i<=n;++i)
    #define PER(i,a,n) for(int i=n;i>=a;--i)
    #define hr putchar(10)
    #define pb push_back
    #define lc (o<<1)
    #define rc (lc|1)
    #define mid ((l+r)>>1)
    #define ls lc,l,mid
    #define rs rc,mid+1,r
    #define x first
    #define y second
    #define io std::ios::sync_with_stdio(false)
    #define endl '
    '
    #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
    using namespace std;
    typedef long long ll;
    typedef pair<int,int> pii;
    const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
    ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
    ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
    ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
    inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
    //head
    
    
    
    const int N = 1e5+10;
    int n, m, lst, s[12][N];
    int pos[12][N];
    int bas[N], has[12][N];
    int Hash(int id, int l, int r) {
    	int ret = (has[id][r]-(ll)has[id][l-1]*bas[r-l+1])%P;
    	if (ret<0) ret += P;
    	return ret;
    }
    int calc(int x, int p1, int y, int p2) {
    	int r=min({lst+1,p1,p2});
    	return Hash(x,p1-r+1,p1)==Hash(y,p2-r+1,p2)?r:1;
    }
    int main() {
    	scanf("%d%d", &n, &m);
    	bas[0] = 1, bas[1] = n*20+1;
    	REP(i,2,n) bas[i]=(ll)bas[i-1]*bas[1]%P;
    	REP(i,1,m) REP(j,1,n) { 
    		scanf("%d", s[i]+j);
    		has[i][j] = ((ll)has[i][j-1]*bas[1]+s[i][j])%P;
    		pos[i][s[i][j]] = j;
    	}
    	if (m==1) return printf("%lld
    ",(ll)n*(n+1)/2),0;
    	ll ans = 0;
    	REP(i,1,n) {
    		int d = 1e9;
    		REP(j,2,m) d = min(d, calc(1,i,j,pos[j][s[1][i]]));
    		ans += d;
    		lst = d;
    	}
    	printf("%lld
    ", ans);
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/10914033.html
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