CSU 1510 Happy Robot

1510: Happy Robot

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 19  Solved: 7

Description

CSU 1510 Happy Robot_第1张图片

 

Input

There will be at most 1000 test cases. Each case contains a command sequence with no more than 1000 characters.

 

Output

For each test case, print the case number, followed by minimal/maximal possible x (in this order), then the minimal/maximal possible y.

 

Sample Input

F?F
L??
LFFFRF

Sample Output

Case 1: 1 3 -1 1
Case 2: -1 1 0 2
Case 3: 1 1 3 3

HINT

 

Source

湖南省第十届大学生计算机程序设计竞赛

 

解题:dp啦,现场居然没做出来。。。笨得还可以。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 1010;
18 char cmd[maxn];
19 int dp[2][4];
20 const int dir[4][2] = {1,0,0,-1,-1,0,0,1};//右 下 左 上
21 int mymax(int a,int b){
22     return max(a,b);
23 }
24 int mymin(int a,int b){
25     return min(a,b);
26 }
27 int go(int (*op)(int,int),bool flag,int x,int y){
28     int cur = 0;
29         dp[0][0] = 0;
30         for(int i = 1; i < 4; ++i)
31             if(flag) dp[0][i] = -INF;
32             else dp[0][i] = INF;
33         for(int i = 0; cmd[i]; ++i) {
34             for(int k = 0; k < 4; ++k) dp[cur^1][k] = flag?-INF:INF;
35             if(cmd[i] == 'F' || cmd[i] == '?') {
36                 dp[cur^1][0] = op(dp[cur^1][0],dp[cur][0]+x);
37                 dp[cur^1][1] = op(dp[cur^1][1],dp[cur][1]-y);
38                 dp[cur^1][2] = op(dp[cur^1][2],dp[cur][2]-x);
39                 dp[cur^1][3] = op(dp[cur^1][3],dp[cur][3]+y);
40             }
41             if(cmd[i] == 'L' || cmd[i] == '?') {
42                 dp[cur^1][0] = op(dp[cur^1][0],dp[cur][1]);
43                 dp[cur^1][1] = op(dp[cur^1][1],dp[cur][2]);
44                 dp[cur^1][2] = op(dp[cur^1][2],dp[cur][3]);
45                 dp[cur^1][3] = op(dp[cur^1][3],dp[cur][0]);
46             }
47             if(cmd[i] == 'R' || cmd[i] == '?') {
48                 dp[cur^1][0] = op(dp[cur^1][0],dp[cur][3]);
49                 dp[cur^1][1] = op(dp[cur^1][1],dp[cur][0]);
50                 dp[cur^1][2] = op(dp[cur^1][2],dp[cur][1]);
51                 dp[cur^1][3] = op(dp[cur^1][3],dp[cur][2]);
52             }
53             cur ^= 1;
54         }
55         int ans = flag?-INF:INF;
56         for(int i = 0; i < 4; ++i)
57             ans = op(dp[cur][i],ans);
58         return ans;
59 }
60 int main() {
61     int cs = 1;
62     while(~scanf("%s",cmd)) {
63         int max_x = go(mymax,true,1,0);
64         int min_x = go(mymin,false,1,0);
65         int max_y = go(mymax,true,0,1);
66         int min_y = go(mymin,false,0,1);
67         printf("Case %d: %d %d %d %d\n",cs++,min_x,max_x,min_y,max_y);
68     }
69     return 0;
70 }
View Code

 

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