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  • CodeForcesGym 100735H Words from cubes

    Words from cubes

    Time Limit: Unknown ms
    Memory Limit: 65536KB
    This problem will be judged on CodeForcesGym. Original ID: 100735H
    64-bit integer IO format: %I64d      Java class name: (Any)
     

    Informikas was cleaning his drawers while he found a toy of his childhood. Well, it's not just a toy, it's a bunch of cubes with letters and digits written on them.

    Informikas remembers that he could have make any word he could think of using these cubes. He is not sure about that now, because some of the cubes have been lost.

    Informikas has already come up with a word he would like to make. Could you help him by saying if the word can be built from the cubes in the drawer?

     

    Input

    On the first line of input there is a string S, consisting of lowercase English letters, and an integer N (4 ≤ |S| ≤ 20, 1 ≤ N ≤ 100) – the word Informikas want to build and the number of cubes. On the every of the following N lines there are 6 characters. Every of those characters is either a lowercase English letter or a digit.

    It is guaranteed that the string S consists only of lowercase English letters.

     

    Output

    Output one word, either "YES", if the word can be built using given cubes, or "NO" otherwise.

     

    Sample Input

    Input
    dogs 4
    d 1 w e 7 9
    o 2 h a v e
    g 3 c o o k
    s 3 i e s 5
    Output
    YES
    Input
    banana 6
    b a 7 8 9 1
    n 1 7 7 7 6
    a 9 6 3 7 8
    n 8 2 4 7 9
    a 7 8 9 1 3
    s 7 1 1 2 7
    Output
    NO

    Source

     
    解题:最大匹配,匈牙利算法
     
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 256;
     4 char str[maxn];
     5 int n,Link[maxn];
     6 bool w[maxn][maxn],used[maxn];
     7 bool match(int u){
     8     for(int i = 0; i < n; ++i){
     9         if(used[i] || !w[u][i]) continue;
    10         used[i] = true;
    11         if(Link[i] == -1 || match(Link[i])){
    12             Link[i] = u;
    13             return true;
    14         }
    15     }
    16     return false;
    17 }
    18 int main(){
    19     while(~scanf("%s",str)){
    20         scanf("%d",&n);
    21         memset(w,false,sizeof w);
    22         char tmp[20];
    23         for(int i = 0; i < n; ++i){
    24             for(int j = 0; j < 6; ++j){
    25                 scanf("%s",tmp);
    26                 for(int k = 0; str[k]; ++k)
    27                     if(str[k] == tmp[0]) w[k][i] = true;
    28             }
    29         }
    30         memset(Link,-1,sizeof Link);
    31         int ret = 0;
    32         for(int i = 0; i < 26; ++i){
    33             memset(used,false,sizeof used);
    34             if(match(i)) ++ret;
    35         }
    36         printf("%s
    ",strlen(str) == ret?"YES":"NO");
    37     }
    38     return 0;
    39 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4812524.html
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