UVA 111 History Grading

Posted 2020-09-22

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=47

LCS类型的题目。

题目中有一点真的读题的时候需要注意：Given the correct chronological order of n events 1; 2; : : : ; n as c1; c2; : : : cn where 1 <= ci <= n denotes the ranking of event i in the correct chronological order .所以 2 3 1 4  指的是event 1 应该被排在第2位，event 2 应该被排在第3位.......

代码如下：

 1 #include <iostream>
 2 #include <math.h>
 3 #include <stdio.h>
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <string.h>
 7 #include <string>
 8 #include <sstream>
 9 #include <cstring>
10 #include <queue>
11 #include <vector>
12 #include <functional>
13 #include <cmath>
14 #define SCF(a) scanf("%d", &a)
15 #define IN(a) cin>>a
16 #define FOR(i, a, b) for(int i=a;i<b;i++)
17 typedef long long Int;
18 using namespace std;
19 
20 int main()
21 {
22     int n = 0;
23     int correct[25];
24     int current[25];
25     int dp[25][25];
26     string in;
27     int state = 2;
28     int cnum;
29     while (getline(cin, in))
30     {
31         stringstream iss;
32         iss << in;
33         if (state == 1)
34         {
35             FOR(i, 0, n)
36             {
37                 iss >> cnum;
38                 correct[cnum-1] = i+1;
39             }
40             state = 2;
41         }
42         else if (state == 2)
43         {
44             int num = 0;
45             while (iss >> cnum)
46             {
47                 current[cnum - 1] = num + 1;
48                 num++;
49             }
50             if (num > 1)
51             {
52                 FOR(i, 0, n)
53                 {
54                     dp[0][i] = dp[i][0] = 0;
55                 }
56                 FOR(i, 1, n + 1)
57                 {
58                     FOR(j, 1, n + 1)
59                     {
60                         if (correct[i - 1] == current[j - 1])
61                             dp[i][j] = dp[i - 1][j - 1] + 1;
62                         else
63                             dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
64                     }
65                 }
66                 printf("%d\n", dp[n][n]);
67             }
68             else
69             {
70                 state = 1;
71                 n = cnum;
72             }
73         }
74     }
75     return 0;
76 }