POJ 3691 DNA repair

POJ_3691

    这个题目可以直接在生成的trie图的“安全部分”进行dp，用f[i][j]表示第i步走到第j个节点时，最少repair了几个基因。

    具体的建立trie图的思想可以参考《Trie图的构建、活用与改进》这篇文章。

#include<stdio.h>
#include<string.h>
#define MAXD 1010
#define INF 0x3f3f3f3f
char gene[MAXD], b[30];
int f[MAXD][MAXD], next[MAXD][4], flag[MAXD], N, e, P[MAXD], q[MAXD];
void add(int cur, int k)
{
    ++ e;
    flag[e] = 0;
    memset(next[e], 0, sizeof(next[e]));
    next[cur][k] = e;
}
void init()
{
    int i, j, k, cur;
    e = 0;
    flag[e] = 0;
    memset(next[e], 0, sizeof(next[e]));
    for(i = 0; i < N; i ++)
    {
        scanf("%s", b);
        cur = 0;
        for(j = 0; b[j]; j ++)
        {
            switch(b[j])
            {
                case 'A' : {k = 0; break;}
                case 'T' : {k = 1; break;}
                case 'C' : {k = 2; break;}
                case 'G' : {k = 3; break;}
            }
            if(!next[cur][k])
                add(cur, k);
            cur = next[cur][k];
        }
        flag[cur] = 1;
    }
    for(i = 0; i < 4; i ++)
        if(!next[0][i])
            add(0, i);
    scanf("%s", gene);
}
void solve()
{
    int i, j, k, u, v, front, rear, ans;
    front = rear = P[0] = 0;
    q[rear ++] = 0;
    while(front < rear)
    {
        u = q[front ++];
        for(i = 0; i < 4; i ++)
        {
            j = next[u][i];
            if(j != 0)
            {
                q[rear ++] = j;
                if(u == 0)
                    P[j] = 0;
                else
                {
                    for(k = P[u]; k != 0; k = P[k])
                        if(next[k][i])
                            break;
                    P[j] = next[k][i];
                    if(flag[P[j]])
                        flag[j] = 1;
                }
            }
            else
            {
                for(k = P[u]; k != 0; k = P[k])
                    if(next[k][i])
                        break;
                next[u][i] = next[k][i];
            }
        }
    }
    memset(f, 0x3f, sizeof(f));
    f[0][0] = 0;
    for(i = 0; gene[i]; i ++)
    {
        switch(gene[i])
        {
            case 'A' : {k = 0; break;}
            case 'T' : {k = 1; break;}
            case 'C' : {k = 2; break;}
            case 'G' : {k = 3; break;}
        }
        for(u = 0; u <= e; u ++)
            if(f[i][u] != INF)
            {
                for(j = 0; j < 4; j ++)
                {
                    v = next[u][j];
                    if(!flag[v])
                    {
                        if(j == k)
                        {
                            if(f[i][u] < f[i + 1][v])
                                f[i + 1][v] = f[i][u];
                        }
                        else
                        {
                            if(f[i][u] + 1 < f[i + 1][v])
                                f[i + 1][v] = f[i][u] + 1;
                        }
                    }
                }
            }
    }
    ans = INF;
    for(j = 0; j <= e; j ++)
        if(f[i][j] < ans)
            ans = f[i][j];
    printf("%d\n", ans == INF ? -1 : ans);
}
int main()
{
    int t = 0;
    for(;;)
    {
        scanf("%d", &N);
        if(!N)
            break;
        printf("Case %d: ", ++ t);
        init();
        solve();
    }
    return 0;
}