[Offer收割]编程练习赛36

逃离单身节

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using std::vector;
using std::sort;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
#define pk putchar(' ')
#define p_ putchar('_')
#define pl putchar('
');
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) pk;
    for (int i = 0; i < n; i++) p_;
    for (int i = 0; i < 3 * n; i++) pk;
    for (int i = 0; i < n; i++) p_;
    pl;
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - 1 - i; j++) pk;
        putchar('/');
        for (int j = 0; j < n + i + i; j++) pk;
        putchar('\');
        for (int j = 0; j < n + n - 1 - i + n - 1 - i; j++) pk;
        putchar('/');
        for (int j = 0; j < n + i + i; j++) pk;
        putchar('\');
        pl;
    }
    putchar('/');
    for (int j = 0; j < 3 * n - 2; j++) pk;
    putchar('\');
    for (int j = 0; j < n; j++) p_;
    putchar('/');
    for (int j = 0; j < 3 * n - 2; j++) pk;
    putchar('\');
    pl;
    for (int i = 0; i < n; i++) {
        putchar('|');
        for (int j = 0; j < 7 * n - 2; j++) pk;
        putchar('|');
        pl;
    }
    for (int i = 0; i < 3 * n; i++) {
        for (int j = 0; j < i; j++) pk;
        putchar('\');
        if (i == 3 * n - 1) for (int j = 0; j < n; j++) p_;
        else for (int j = 0; j < 7 * n - 2 - 2 * i; j++) pk;
        putchar('/');
        pl;
    }
    return 0;
}

 小Hi的天平

在并查集和二分图之间纠结了好久，直到看了别人的代码才想明白。一开始想的并查集是如果两件物品重量相等就合并两个集合，但这样一来面对一条不等关系的信息就毫无作为了。其实相等和不等表达的信息是等价的，都可以由第一个的种类推出第二个。这样一来，在同一个集合中的物品就之间的关系就不是“属于同一种物品”，而是“他们之间有逻辑关系”，只要确定一个集合中的任意一个是属于A还是B就能确定集合中所有的物品的种类。所以，读入一条xuv后，如果uv属于同一个集合，只需核实其关系是否正确即可；若不属于同一个集合，就合并两个集合。这题的关键代码是写在并查集的路径压缩上，一开始隐隐约约有感觉，但是太菜了，没想到。

c[x]表示x节点与其所在集合根节点是否相同

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using std::vector;
using std::sort;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
class Union_Find_Set {
#define MAX_UNION_FIND_SET_SIZE 100000
public:
    int setSize;
    int father[MAX_UNION_FIND_SET_SIZE];
    int c[MAX_UNION_FIND_SET_SIZE];
    Union_Find_Set() {
        setSize = 0;
    }
    Union_Find_Set(int x) {
        setSize = x;
        clear();
    }
    void clear() {
        for (int i = 0; i < setSize; i++) {
            father[i] = i;
        }
        memset(c, 0, sizeof(c));
    }
    int getFather(int x) {
        if (x == father[x]) return x;
        int y = father[x];
        father[x] = getFather(father[x]);
        c[x] = c[x] ^ c[y];
        return father[x];
    }
    bool merge(int x, int a, int b) {
        int fa = getFather(a), fb = getFather(b);
        if (fa != fb) {
            father[fa] = fb;
            c[fa] = x ^ c[a] ^ c[b];
            return true;
        } else {
            return false;
        }
    }
    int countRoot() {
        int ret = 0;
        for (int i = 0; i < setSize; i++) {
            if (father[i] = i) {
                ret++;
            }
        }
        return ret;
    }
};
Union_Find_Set ufs(10005);
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, m, t;
    bool flag;
    scanf("%d", &t);
    while (t--) {
        flag = true;
        scanf("%d%d", &n, &m);
        ufs.clear();
        for (int i = 1; i <= m; i++) {
            int x, u, v;
            scanf("%d%d%d", &x, &u, &v);
            if (!flag) continue;
            int fu = ufs.getFather(u), fv = ufs.getFather(v);
            if (fu == fv) {
                if ((ufs.c[u] ^ ufs.c[v]) != (x)) {
                    printf("sad
%d
", i);
                    flag = false;
                }
            } else {
                ufs.merge(x, u, v);
            }
        }
        if (flag) printf("great
");
    }
    return 0;
}

 图书馆

剩下30%的数据实在无能为力，绝望。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
long long a[1005];
bool f[22005];
int p[10000], cnt, c[10000];
vector<int> q[10005], qq[10005];
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    memset(f, true, sizeof(f));
    f[0] = f[1] = false;
    for (int i = 2; i < 10000; i++)
        for (int j = i + i; j < 10000; j += i)
            f[j] = false;
    cnt = 0;
    for (int i = 0; i < 10000; i++) if (f[i]) p[cnt++] = i;
    for (int i = 0; i <= 10000; i++) {
        q[i].clear();
        qq[i].clear();
    }
    for (int i = 2; i <= 10000; i++) {
        int x = i;
        for (int j = 0; j < cnt; j++) {
            if (x == 1) break;
            int num = 0;
            while (x % p[j] == 0) {
                num++;
                x /= p[j];
            }
            if (num > 0) {
                q[i].push_back(j);
                qq[i].push_back(num);
            }
        }
    }
    int n, t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            scanf("%lld", &a[i]);
            if (a[i] > 100) a[i] %= 100;
        }
        if (n > 100) {
            printf("0
");
            continue;
        }
        memset(c, 0, sizeof(c));
        int sum = 0;
        for (int i = 0; i < n; i++) sum += (int)a[i];
        for (int i = 2; i <= sum; i++) {
            for (int j = 0; j < q[i].size(); j++) {
                c[q[i][j]] += qq[i][j];
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 2; j <= a[i]; j++) {
                for (int k = 0; k < q[j].size(); k++) {
                    c[q[j][k]] -= qq[j][k];
                }
            }
        }
        int ans = 1;
        for (int i = 0; i < cnt; i++) {
            for (int j = 0; j < c[i]; j++) {
                ans = (ans * p[i]) % 10;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}

 命名的烦恼

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
using std::vector;
using std::queue;
using std::map;
using std::sort;
using std::string;
#define read(x) scanf("%d", &x)
#define reads(x) scanf("%s", x)
#define write(x) printf("%d ", x)
#define writes(x) printf("%s ", x)
#define writeln(x) printf("%d
", x)
#define writesln(x) printf("%s
", x)

int cmp(const void * x, const void * y) {
#define datatype int
    datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
    //x < y
    return dx > dy ? 1 : -1;
#undef datatype
}

char a[20][105], str[20][105];
int cost[20][20], dp[1 << 16][20];
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, m = 0;
    read(n);
    for (int i = 0; i < n; i++) reads(a[i]);
    for (int i = 0; i < n; i++) {
        bool flag = false;
        for (int j = 0; j < n; j++) {
            if (i == j) continue;
            if (flag) break;
            int li = strlen(a[i]), lj = strlen(a[j]);
            for (int k = 0; k + li - 1 < lj; k++) {
                bool find = true;
                for (int p = 0; p < li; p++) {
                    if (a[i][p] != a[j][k + p]) {
                        find = false;
                        break;
                    }
                }
                if (find) {
                    flag = true;
                    break;
                }
            }
        }
        if (!flag) {
            memcpy(str[m], a[i], sizeof(a[i]));
            m++;
        }
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < m; j++) {
            if (i == j) continue;
            int common, li = strlen(str[i]), lj = strlen(str[j]);
            for (common = 0; common < li && common < lj; common++) {
                bool match = true;
                for (int k = 0; k < common; k++) {
                    if (str[i][li - common + k] != str[j][k]) {
                        match = false;
                        break;
                    }
                }
                if (match) cost[i][j] = lj - common;
            }
        }
    }
    memset(dp, 0x3F, sizeof(dp));
    for (int i = 0; i < m; i++) {
        dp[1 << i][i] = strlen(str[i]);
    }
    for (int i = 0; i < (1 << m); i++) {
        for (int j = 0; j < m; j++) {
            if (dp[i][j] == 0x3F3F3F3F) continue;
            for (int k = 0; k < m; k++) {
                if (i & (1 << k)) continue;
                if (dp[i | (1 << k)][k] > dp[i][j] + cost[j][k])
                    dp[i | (1 << k)][k] = dp[i][j] + cost[j][k];
            }
        }
    }
    int ans = 0x3FFFFFFF;
    for (int i = 0; i < m; i++) {
        if (dp[(1 << m) - 1][i] < ans) ans = dp[(1 << m) - 1][i];
    }
    writeln(ans);
    return 0;
}