zoj 1102 Phylogenetic Trees Inherited

这题刚开始的时候没读懂题目，不知道怎么计算ancester。 后来在网上搜了别人的代码，用位操作与或和移位来实现的，

没看太明白里面的移位的规律。后来自己又想了想，感觉应该是集合运算，如果有相同的char就求交集，否则就求并集。

求并集的话肯定有一条边的cost要+1。记着每个序列按字符猪哥计算ancester，否则中间anc数组要设置为[2058][1010][20].

一遍就AC了, 思路正确，最后代码如下。

 

#include <stdio.h>
#include <string.h>

int intersect(char *src1, char *src2, char *dest)
{
        int i, j, len1, len2, k;
        len1 = strlen(src1), len2 = strlen(src2);
        k = 0;
        for(i=0;i<len1;i++) {
                for(j=0;j<len2;j++) {
                        if (src2[j] == src1[i]) {
                                dest[k++] = src1[i];
                                break;
                        }
                }
        }
        return k;
}

int unionsect(char *src1, char *src2, char *dest)
{
        int i, j, len = strlen(src2);
        char *cptr;
        j = strlen(src1);
        strncpy(dest, src1,j);
        for(i=0;i<len;i++) {
                cptr = strchr(dest, src2[i]);
                if (!cptr) {
                        dest[j++] = src2[i];
                }
        }
        return j;
}

int func(char *src1, char *src2, char *dest)
{
        if (!intersect(src1, src2, dest)) {
                unionsect(src1, src2, dest);
                return 1;
        }
        return 0;
}

int main()
{
#ifdef LOCAL_INPUT
        freopen("data", "r", stdin);
#endif
        int n, l, i, j, cost;
        char seq[1030][1010];
        char ind[2060][20];

        while(scanf("%d %d", &n, &l) &&(n || l)) {
                for(i=0;i<n;i++) {
                        scanf("%s", seq[i]);
                }
                cost = 0;
                for(i=0;i<l;i++) {
                        memset(ind, '\0', sizeof(ind));
                        for(j=0;j<n;j++) {
                                ind[j+n][0] = seq[j][i];
                        }
                        for(j=n-1;j>0;j--) {
                                cost += func(ind[j*2], ind[j*2+1], ind[j]);
                        }
                        printf("%c", ind[1][0]);
                }
                printf(" %d\n", cost);
        }
        return 0;
}