UVa 10618 Tango Tango Insurrection

题目：

Problem A: Tango Tango Insurrection

You are attempting to learn to play a simple arcade dancing game. The game has 4 arrows set into a pad: Up, Left, Down, Right. While a song plays, you watch arrows rise on a screen, and when they hit the top, you have to hit the corresponding arrows on the pad. There is no penalty for stepping on an arrow without need, but note that merely standing on an arrow does not activate it; you must actually tap it with your foot. Many sequences in the game are very fast-paced, and require proper footwork if you don't want to tire yourself out. Write a program to determine the easiest way to execute a certain sequence of arrows.

We will work with a basic time unit of an eighth-note. At any given time, your left foot and right foot will each be on distinct arrows. Only one foot may perform an action (changing arrows and/or tapping) during any time unit; jumping is not allowed. Also, you must remain facing forward in order to see the screen. This puts limitations on which feet you can use to hit which arrows. Finally, hitting two arrows in a row with the same foot ("double-tapping") is exhausting, because you can't shift your weight onto that foot. Ideally, you want to alternate feet all the way through a string of consecutive arrows.

Performing an action with a foot costs 1 unit of energy if it did NOT perform an action in the previous time unit. If it did, then it costs 3 units if it doesn't change arrows, 5 units if it moves to an adjacent arrow, and 7 units if it moves directly across the pad (between Up and Down, or Left and Right).

Under normal circumstances, you can't put your left foot on Right, or your right foot on Left. However, you CAN do a temporary "crossover": if your left foot is on Up or Down, you can twist your hips and put your right foot on Left - but until your right foot moves away, you can't move your left to a different arrow. (Imagine the tangle your legs would get into if you tried!) Similarly, you can cross your left foot over/behind your right.

Input

You will be given multiple arrow sequences to provide foot guides for. Every sequence consists of a line containing from 1 to 70 characters, representing the arrow that must be hit at each time unit. The possible characters are U, L, D, and R, signifying the four arrows, or a period, indicating that no arrow need be hit. Assume that your left and right feet start on the Left and Right arrows for the first time unit of a sequence.

There are at most 100 sequences. Input is terminated by a line consisting of a single #.

Output

For each input sequence, output a string of the same length, indicating which foot should perform an action at each time step, or '.' if neither does. If there are multiple solutions that require minimal energy, any will do.

Sample Input

LRLRLLLLRLRLRRRRLLRRLRLDU...D...UUUUDDDD

#

Possible Output for Sample Input

LRLRLLLLRLRLRRRRLLRRLRLRL...R...LLLLRRRR

----------------------------------------------------------------------------------------------------------------------------------------------

思路：

  对于屏幕上的位置必须有一脚踩下，对两脚位置有所要求且根据脚的移动关系分配代价，求完成屏幕要求的情况下代价最小。

用状态d[i][a][b][s]表示已踩过i个命令，左右脚位置为ab，因为需要根据当前移动的脚是否刚移动过所以用s表示上次移动的脚。

状态转移方程： d[i][a][b][s]=min(d[i][ta][tb][s']+cost)

  但注意到，expr是当前的移动，移动后转移到i+1且位置成为移动后的位置， 因此需要倒序枚举i，把i+1看作是 i 的子问题

原来char[]可以这么用。

代码：

 #include<cstdio>

 #include<cassert>

 #include<cstring>

 #define FOR(a,b,c) for(int a=(b);a<(c);a++)

 const int maxn =  + ;

 const int UP =,LEFT=,RIGHT=,DOWN=;

 //const direction

 struct Node{

     int f,t;

 };

 char expr[maxn],footch[]=".LR";

 char pos[];   //char pos[char]=int

 int d[maxn][][][];

 Node action[maxn][][][];

 int energy(int a,int ta) {

     if(a==ta) return ;   //没有移动

     if(a + ta == ) return ;  //相对箭头 03 12

     return ;   //else

 }

 int energy(int i,int a,int b,int s,int f,int t,int& ta,int& tb) {

     ta=a; tb=b;

     if(f==LEFT) ta=t;

     else if(f==RIGHT) tb=t;

     if(ta==tb) return -;

     if(ta==RIGHT && tb==LEFT) return -;

     if(a==RIGHT && tb!=b)   return -; //左脚在右 此时右脚不能动

     if(b==LEFT  && ta!=a)    return -;//右脚在左 此时左脚不能动 

     if(f==) return ;      //没有移动

     else  if(f != s) return ;  //当前移动的脚上一次没有动

     else {   //动了

         if(f==) return energy(a,ta); //当前动的左脚

         else return energy(b,tb);    //右脚

     }

 }

 void update (int i,int a,int b,int s,int f,int t) {

 //状态(i,a,b,s) 将脚f移动到位置t

     int ta,tb;

     int v=energy(i,a,b,s,f,t,ta,tb);

     if(v<) return;  //移动不合法 

     int& ans=d[i][a][b][s];

     //? : 因为s代表的是前一个移动的脚 倒叙枚举方便识别子问题

     int cost=v+d[i+][ta][tb][f];

     if(cost < ans) {

         ans=cost;

         action[i][a][b][s]=(Node){f,t};

     }

 } 

 int main(){

     pos['U']=; pos['L']=; pos['R']=; pos['D']=;

     while(scanf("%s",expr)== && expr[]!='#') {

         int n=strlen(expr);

         memset(d,,sizeof(d));

         //为什么i要倒序枚举 ?

         //state(i,a,b,s) 代表已经执行i-尾  左右脚为ab 前一次移动了s脚

         for(int i=n-;i>=;i--)

           FOR(a,,)

             FOR(b,,) if(a!=b)

               FOR(s,,) {

                 d[i][a][b][s]=*n;

                   if(expr[i]=='.'){

                       update(i,a,b,s,,); //不移动

                       FOR(t,,) {          // |随便移动

                           update(i,a,b,s,LEFT,t);

                           update(i,a,b,s,RIGHT,t);

                       }

                   }

                   else

                    update(i,a,b,s,LEFT,pos[expr[i]]);

                    update(i,a,b,s,RIGHT,pos[expr[i]]);

               }

         int a=LEFT,b=RIGHT,s=;

         FOR(i,,n) {  //output 动哪一只脚的序列

             int f=action[i][a][b][s].f;

             int t=action[i][a][b][s].t;

             printf("%c",footch[f]);

             s=f;

             if(f==) a=t;

             else if(f==) b=t;

             //f==0 ab不动

         }

         printf("\n");

     }

     return ;

 }