【HDOJ】2371 Decode the Strings

快速矩阵乘法。注意，原始字符串即为decode后的字符串。题目是要找到原始串。

#include <cstdio>
#include <cstring>

#define MAXN 85

typedef struct {
    char m[MAXN][MAXN];
} mat_st;

int n, m;
char buf[MAXN];
mat_st e;

mat_st mat_mult(mat_st a, mat_st b) {
    int i, j, k;
    mat_st c;
    memset(c.m, 0, sizeof(c.m));

    for (i=1; i<=n; ++i) {
        for (j=1; j<=n; ++j) {
            for (k=1; k<=n; ++k)
                c.m[i][j] += a.m[i][k] & b.m[k][j];
        }
    }
    return c;
}

mat_st mat_power(mat_st a, int r) {
    mat_st ret;

    memset(ret.m, 0, sizeof(ret.m));
    for (int i=1; i<=n; ++i)
        ret.m[i][i] = 1;

    while (r) {
        if (r & 1)
            ret = mat_mult(ret, a);
        a = mat_mult(a, a);
        r >>= 1;
    }
    return ret;
}

int main() {
    mat_st org;
    int i, j;

    while (scanf("%d %d",&n,&m)!=EOF && n) {
        memset(org.m, 0, sizeof(org.m));
        for (i=1; i<=n; ++i) {
            scanf("%d", &j);
            org.m[i][j] = 1;
        }
        getchar();
        gets(buf+1);
        mat_st ans = mat_power(org, m);
        for (i=1; i<=n; ++i) {
            for (j=1; j<=n; ++j) {
                if (ans.m[j][i]) {
                    printf("%c", buf[j]);
                    break;
                }
            }
        }
        printf("
");
    }

    return 0;
}