LightOJ 1013 - Love Calculator

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题意:

题意就是给你两个字符串,求能同事包含这两个字符串的最短串的长度和种数。一个结论:最长包含串的长度=len(s)+len(t)-len(lcs(s,t)).另外统计的方法和求lcs的方程转移差不多。
////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define ALL(x) x.begin(),x.end()#define AT(i,v) for (auto &i:v)#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define look(x) cout << #x << "=" << x#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bignconst int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;const int N=40;int len[N][N];ll dp[N][N];int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    int T_T;cin >> T_T;    getchar();    for (int kase=1;kase<=T_T;kase++) {        char s[N],t[N];        gets(s+1);gets(t+1);//      puts(s+1);puts(t+1);        CLR(len);CLR(dp);        int ls=strlen(s+1),lt=strlen(t+1);        for (int i=1;i<=ls;i++) {            for (int j=1;j<=lt;j++) {                if (s[i]==t[j]) len[i][j]=len[i-1][j-1]+1;                else len[i][j]=max(len[i-1][j],len[i][j-1]);            }        }        for (int i=0;i<=ls;i++) dp[i][0]=1;        for (int i=0;i<=lt;i++) dp[0][i]=1;        for (int i=1;i<=ls;i++) {            for (int j=1;j<=lt;j++) {                if (s[i]==t[j]) dp[i][j]+=dp[i-1][j-1];                else {                    if (len[i][j]==len[i-1][j]) dp[i][j]+=dp[i-1][j];                    if (len[i][j]==len[i][j-1]) dp[i][j]+=dp[i][j-1];                }            }        }        printf("Case %d: %d %lld\n",kase,ls+lt-len[ls][lt],dp[ls][lt]);    }    return 0;}
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