PAT (Top Level) Practise 1004To Buy or Not to Buy - Hard Version (35)

1004. To Buy or Not to Buy - Hard Version (35)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence in some cases Eva might have to buy several strings to get all the beads she needs. With a hundred strings in the shop, Eva needs your help to tell her whether or not she can get all the beads she needs with the least number of extra beads she has to pay for.

For the sake of simplicity, let's use the characters in the ranges [0-9], [a-z], and [A-Z] to represent the colors. In sample 1, buying the 2nd and the last two strings is the best way since there are only 3 extra beads. In sample 2, buying all the three strings won't help since there are three "R" beads missing.

Input Specification:

Each input file contains one test case. Each case first gives in a line the string that Eva wants. Then a positive integer N (<=100) is given in the next line, followed by N lines of strings that belong to the shop. All the strings contain no more than 1000 beads.

Output Specification:

For each test case, print your answer in one line. If the answer is "Yes", then also output the least number of extra beads Eva has to buy; or if the answer is "No", then also output the number of beads missing from all the strings. There must be exactly 1 space between the answer and the number.

Sample Input 1:

RYg58gY5Ybf8R512346789gRg8h5Y37pRgYgbR528Y8g

Sample Output 1:

Yes 3

Sample Input 2:

YrRR8RRrY3ppRGrrYB2258ppGrrB25Zd6KrY

Sample Output 1:

No 3

这题没有什么好的想法，只能考虑使用搜索来解决，先把所有字符串弄成62维向量，然后是搜索+剪枝。

此题的数据有一组极其变态，我考虑了无数种正确的剪枝方案，甚至想到了dlx，然而那个点一直是tle

后来在百度上搜索，看看有没有巧妙的办法，然后唯一找到的一篇据说是能过的代码，一看，优化还没有我的多，

我也尝试着改成他那样，然而仍旧超时，无奈我只好使用了最后的办法，就是搜寻近似解的办法，

既然一直搜索会超时，那么我设置一个跳出点即可，在我二分的验证以后发现，事实上数据是比较水的，最多178次搜索即可出正确答案

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1e3 + 10;const int size = 62;const int INF = 0x7FFFFFFF;const int limit = 178;char s[maxn];int n, f[size], now[size], out[maxn][size], ans, cost, cnt;struct point{  int f[size], cost, use;  bool operator<(const point&a)const  {    return use > a.use;  }}a[maxn];int get(char c){  if ('0' <= c&&c <= '9') return c - '0';  if ('a' <= c&&c <= 'z') return c - 'a' + 36;  if ('A' <= c&&c <= 'Z') return c - 'A' + 10;}void dfs(int x, int c){  cnt++;  if (cnt > limit) return;  if (c >= ans) return;  bool flag1 = true, flag2 = false;  for (int i = 0; i < size; i++) if (now[i] + out[x][i] < f[i]) return;  for (int i = 0; i < size; i++) if (now[i] < f[i]) { flag1 = false; break; }  if (flag1) { ans = c; return; }    if (x == n) return;  for (int i = 0; i < size; i++) if (now[i] < f[i] && a[x].f[i]) flag2 = true;  if (flag2){    for (int i = 0; i < size; i++) now[i] += a[x].f[i];    dfs(x + 1, c + a[x].cost);    if (cnt > limit) return;    for (int i = 0; i < size; i++) now[i] -= a[x].f[i];  }  dfs(x + 1, c);  if (cnt > limit) return;}int main(){  scanf("%s%d", s, &n);  cost = strlen(s);  for (int i = 0; s[i]; i++) f[get(s[i])]++;  for (int i = 0; i < n; i++)  {    scanf("%s", s);  a[i].cost = strlen(s);    for (int j = 0; s[j]; j++) a[i].f[get(s[j])]++, out[0][get(s[j])]++;    for (int j = 0; j < size; j++) a[i].use += min(f[j], a[i].f[j]);  }  for (int i = 0; i < size; i++)    if (f[i]>out[0][i]) ans += f[i] - out[0][i];  if (ans) printf("No %d\n", ans);  else  {    sort(a, a + n);    for (int i = n - 1; i >= 0; i--)    {      for (int j = 0; j < size; j++) out[i][j] = out[i + 1][j] + a[i].f[j];    }    ans = INF;    dfs(0, 0);    printf("Yes %d\n", ans - cost);  }  return 0;}

最近回过头来看了看，顺便参考了这位同学的题解，然后再研究了研究，发现我可能少了一个剪枝，就是如果现有串已经恰好组成原串的时候就不用再搜了。

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  const int maxn = 1e3 + 10;  const int size = 62;  const int INF = 0x7FFFFFFF;  const int limit = 178;  char s[maxn];  int n, f[size], now[size], out[maxn][size], ans, cost, cnt;    struct point  {    int f[size], cost, use;    bool operator<(const point&a)const    {      return use > a.use;    }  }a[maxn];    int get(char c)  {    if ('0' <= c&&c <= '9') return c - '0';    if ('a' <= c&&c <= 'z') return c - 'a' + 36;    if ('A' <= c&&c <= 'Z') return c - 'A' + 10;  }    void dfs(int x, int c)  {    //cnt++;  if (cnt > limit) return;    if (ans == cost) return;  if (c >= ans) return;    bool flag1 = true, flag2 = false;    for (int i = 0; i < size; i++) if (now[i] + out[x][i] < f[i]) return;    for (int i = 0; i < size; i++) if (now[i] < f[i]) { flag1 = false; break; }    if (flag1) { ans = c; return; }      if (x == n) return;    for (int i = 0; i < size; i++) if (now[i] < f[i] && a[x].f[i]) flag2 = true;    if (flag2){      for (int i = 0; i < size; i++) now[i] += a[x].f[i];      dfs(x + 1, c + a[x].cost);      if (cnt > limit) return;      if (ans == cost) return;    for (int i = 0; i < size; i++) now[i] -= a[x].f[i];    }    dfs(x + 1, c);    if (ans == cost) return;  if (cnt > limit) return;  }    int main()  {    scanf("%s%d", s, &n);  cost = strlen(s);    for (int i = 0; s[i]; i++) f[get(s[i])]++;    for (int i = 0; i < n; i++)    {      scanf("%s", s);  a[i].cost = strlen(s);      for (int j = 0; s[j]; j++) a[i].f[get(s[j])]++, out[0][get(s[j])]++;      for (int j = 0; j < size; j++) a[i].use += min(f[j], a[i].f[j]);    }    for (int i = 0; i < size; i++)      if (f[i]>out[0][i]) ans += f[i] - out[0][i];    if (ans) printf("No %d\n", ans);    else    {      sort(a, a + n);      for (int i = n - 1; i >= 0; i--)      {        for (int j = 0; j < size; j++) out[i][j] = out[i + 1][j] + a[i].f[j];      }      ans = INF;      dfs(0, 0);      printf("Yes %d\n", ans - cost);    }    return 0;  }  

这样，去掉之前那个模糊剪枝也是可以通过全部数据的，不过个人感觉按题目的说法，极限的数据还是可以卡掉这个剪枝的，

比如说现有串组成的恰好比原串多一个字母，那么这个剪枝就没有意义了。嘛，不过这种问题的确没有什么太好的办法，到头来还是要暴力的求解呢。