poj1699

题意：给定若干基因片段，求包含所有片段的最短基因序列长度。

分析：cost[i][j]表示把j串接在i穿后面最少需要添加几个字符，先求出整个cost矩阵，然后枚举这些串的全排列，对于每种排列求长度，更新最小值。

View Code

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

#define maxn 15
#define maxl 25

int n;
char gene[maxn][maxl];
int cost[maxn][maxn];
int f[maxn];

void input()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%s", gene[i]);
}

int cal(char *st1, char *st2)
{
    int len1 = strlen(st1);
    int len2 = strlen(st2);
    for (int i = len1 - len2; i < len1; i++)
    {
        bool ok = true;
        for (int j = i; j < len1; j++)
            if (st1[j] != st2[j - i])
            {
                ok = false;
                break;
            }
        if (ok)
            return len2 + i - len1;
    }
    return len2;
}

void make()
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            cost[i][j] = cal(gene[i], gene[j]);
}

void work()
{
    for (int i = 0; i < n; i++)
        f[i] = i;
    int ans = 0x3f3f3f3f;
    do
    {
        int temp = strlen(gene[f[0]]);
        for (int i = 0; i < n - 1; i++)
            temp += cost[f[i]][f[i + 1]];
        ans = min(ans, temp);
    } while (next_permutation(f, f + n));
    printf("%d\n", ans);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    while (t--)
    {
        input();
        make();
        work();
    }
    return 0;
}