SOJ-1015

/******************************************************************************************************
 ** Copyright (C) 2011.05.01 - 2013.07.01
 ** Author: famousDT <[email protected]>
 ** Edit date: 2011-04-25
******************************************************************************************************/
#include <stdio.h>
#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll
#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <string.h>//memcpy(to,from,count
#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprint
#include <algorithm>
using namespace std;

//typedef long long int ll;

#define PI acos(-1)
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))
#define FABS(a) ((a) >= 0 ? (a) : (-(a)))

int main()
{    
    int n;
    int t;
    int i, j; 
    int d[30];
    int s[30];
    cin>>n;
    for (i = 0; i < n; ++i) {
        cin>>t;
        d[t] = i + 1;
    }
    int index = 1;    
    while (scanf("%d", &t) == 1) {
        s[t] = index++;
        int l[30][30] = {0};
        for (i = 1; i < n; ++i) {
            scanf("%d", &t);
            s[t] = index++;
        }
        index = 1;
        for (i = 1; i <= n; ++i) {
            for (j = 1; j <= n; ++j) {
                if (d[i] == s[j])
                    l[i][j] = l[i - 1][j - 1] + 1;
                else 
                    l[i][j] = MAX(l[i - 1][j], l[i][j - 1]);
            }
        }
        printf("%d\n", l[n][n]);
    }
    return 0;
}

 

你可能感兴趣的:(c,Date,String,float)