Uva 省赛傻逼题 B

要不要这么坑，自己的程序总是漏洞百出，不是这种情况没写好，就是那种情况没算进去，分析也不能很好的理论分析！

Problem B

Time limit: 1.000 seconds

Boxes in a Line

You have n boxes in a line on the table numbered 1~n from left to right. Your task is to simulate 4 kinds of commands:

1 X Y: move box X to the left to Y (ignore this if X is already the left of Y)
2 X Y: move box X to the right to Y (ignore this if X is already the right of Y)
3 X Y: swap box X and Y
4: reverse the whole line.

Commands are guaranteed to be valid, i.e. X will be not equal to Y.

For example, if n=6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing 2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1. Then after executing 4, then line becomes 1 3 5 4 6 2

Input

There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m(1<=n, m<=100,000). Each of the following m lines contain a command.

Output

For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n from left to right.

Sample Input

Output for the Sample Input

Case 1: 12
Case 2: 9
Case 3: 2500050000

The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include <map>
  3 #include <queue>
  4 #include <vector>
  5 #include <string>
  6 #include <cstdio>
  7 #include <cstring>
  8 #include <iostream>
  9 #include <algorithm>
 10 using namespace std;
 11 #define maxn 1000005
 12 #define ll long long
 13 #define INF 0x7fffffff
 14 int p[2][maxn][2];
 15 int n,m;
 16 void init(){
 17     memset(p,0,sizeof p);
 18     for(int i=1;i<=n;i++){
 19         p[0][i][0]=i-1;
 20         p[0][i][1]=i+1;
 21         p[1][n-i+1][0]=n-i+2;
 22         p[1][n-i+1][1]=n-i;
 23     }
 24     p[0][0][1]=1;
 25     p[0][n+1][0]=n;
 26     p[1][n+1][1]=n;
 27     p[1][0][0]=1;
 28 }
 29  void fun1(int t,int a,int b){
 30     int al=p[t][a][0];
 31     int ar=p[t][a][1];
 32     if(ar==b)return;
 33     p[t][al][1]=ar;
 34     p[t][ar][0]=al;
 35     int bl=p[t][b][0];
 36     p[t][bl][1]=a;
 37     p[t][a][0]=bl;
 38     p[t][a][1]=b;
 39     p[t][b][0]=a;
 40     t^=1;
 41     al=p[t][a][0];
 42     ar=p[t][a][1];
 43     p[t][al][1]=ar;
 44     p[t][ar][0]=al;
 45     int br=p[t][b][1];
 46     p[t][br][0]=a;
 47     p[t][a][1]=br;
 48     p[t][a][0]=b;
 49     p[t][b][1]=a;
 50  }
 51  void fun2(int t,int a,int b){
 52     int al=p[t][a][0];
 53     int ar=p[t][a][1];
 54     if(al==b)return;
 55     p[t][al][1]=ar;
 56     p[t][ar][0]=al;
 57     int br=p[t][b][1];
 58     p[t][br][0]=a;
 59     p[t][a][1]=br;
 60     p[t][a][0]=b;
 61     p[t][b][1]=a;
 62     t^=1;
 63     al=p[t][a][0];
 64     ar=p[t][a][1];
 65     p[t][al][1]=ar;
 66     p[t][ar][0]=al;
 67     int bl=p[t][b][0];
 68     p[t][bl][1]=a;
 69     p[t][a][0]=bl;
 70     p[t][a][1]=b;
 71     p[t][b][0]=a;
 72  }
 73  void fun3(int t,int a,int b){
 74     int al=p[t][a][0];
 75     int ar=p[t][a][1];
 76     int bl=p[t][b][0];
 77     int br=p[t][b][1];
 78     if(ar!=b&&al!=b){
 79     p[t][al][1]=b;
 80     p[t][b][0]=al;
 81     p[t][ar][0]=b;
 82     p[t][b][1]=ar;
 83     p[t][bl][1]=a;
 84     p[t][a][0]=bl;
 85     p[t][br][0]=a;
 86     p[t][a][1]=br;
 87     t^=1;
 88     al=p[t][a][0];
 89     ar=p[t][a][1];
 90     bl=p[t][b][0];
 91     br=p[t][b][1];
 92     p[t][al][1]=b;
 93     p[t][b][0]=al;
 94     p[t][ar][0]=b;
 95     p[t][b][1]=ar;
 96     p[t][bl][1]=a;
 97     p[t][a][0]=bl;
 98     p[t][br][0]=a;
 99     p[t][a][1]=br;
100     return;
101     }
102     if(ar==b){
103         p[t][al][1]=b;
104         p[t][b][0]=al;
105         p[t][a][0]=b;
106         p[t][b][1]=a;
107         p[t][br][0]=a;
108         p[t][a][1]=br;
109         t^=1;
110         al=p[t][a][0];
111         ar=p[t][a][1];
112         bl=p[t][b][0];
113         br=p[t][b][1];
114         p[t][bl][1]=a;
115         p[t][a][0]=bl;
116         p[t][a][1]=b;
117         p[t][b][0]=a;
118         p[t][ar][0]=b;
119         p[t][b][1]=ar;
120         return;
121     }
122     if(al==b){
123         p[t][bl][1]=a;
124         p[t][a][0]=bl;
125         p[t][b][0]=a;
126         p[t][a][1]=b;
127         p[t][ar][0]=b;
128         p[t][b][1]=ar;
129         t^=1;
130         al=p[t][a][0];
131         ar=p[t][a][1];
132         bl=p[t][b][0];
133         br=p[t][b][1];
134         p[t][al][1]=b;
135         p[t][b][0]=al;
136         p[t][a][0]=b;
137         p[t][b][1]=a;
138         p[t][br][0]=a;
139         p[t][a][1]=br;
140     }
141  }
142  ll sum(int t){
143     ll ans=0;
144     ll x=0;
145     if(t==0)x=0;
146     else x=n+1;
147     for(int i=0;i<n;i++){
148         x=p[t][x][1];
149         if(i%2==0)ans+=x;
150     }
151     return ans;
152  }
153 int main(){
154     int cas=1;
155     while(~scanf("%d%d",&n,&m)){
156         init();
157         int t=0;
158         for(int i=0;i<m;i++){
159             int op,a,b;
160             scanf("%d",&op);
161             if(op==4){t^=1;continue;}
162             scanf("%d%d",&a,&b);
163             if(op==1)fun1(t,a,b);
164             if(op==2)fun2(t,a,b);
165             if(op==3&&a!=b)fun3(t,a,b);
166         }
167         printf("Case %d: ",cas++);
168         printf("%lld\n",sum(t));
169     }
170     return 0;
171 }

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posted @ 2013-10-16 11:53 HaibaraAi 阅读(117) 评论(0) 收藏举报

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