uva 910 - TV Game

url: http://uva.onlinejudge.org/external/9/910.html

helpful discussion: http://acm.uva.es/board/viewtopic.php?f=34&t=16714&p=61697&hilit=910&sid=5e08f43bf5aa383425db1a424d7b794d#p61697

helpful test Case:

5
A B E x
B D C -
C D A -
D D B -
E E E -
5
5
A B E -
B D C x
C D A -
D D B -
E E E -
5
5
A B E -
B D C -
C D A x
D D B -
E E E -
5
5
A B E -
B D C -
C D A -
D D B x
E E E -
5
5
A B E -
B D C -
C D A -
D D B -
E E E x
5
5
A B E x
B D C x
C D A x
D D B x
E E E x
5

desired output:

1
3
3
7
18
32

AC code:

/*
Problem ID: 910
Solver: songyy
Start Time:
End Time:
Time Spent:
*/

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string.h>
#include <string>

usingnamespace std;

int edge0[30], edge1[30], isDes[30];
int memo[30][31];

int cal(int u,int k){
    if(memo[u][k]!=-1)    return memo[u][k];
    
    if(k==0){
         if(isDes[u]){
            memo[u][k] =1;
        } else memo[u][k] =0;
        return memo[u][k];
    }
    
    memo[u][k] = cal(edge0[u],k-1) + cal(edge1[u],k-1);
    
    return memo[u][k];
}

int main(){
    
//    freopen("910_in.txt","r",stdin);
    
    int N;
    char c0,c1,isD;
    int Q;
    
    while(scanf("%d%*c",&N)!=-1){
        memset(edge0,0,sizeof(edge0));
        memset(edge1,0,sizeof(edge1));
        memset(isDes,0,sizeof(isDes));
            
        for(int i=0;i<30;i++){
            for(int j=0;j<31;j++){
                memo[i][j] =-1;
            }
        }
        //memo[0][0] = 1;
        
        //printf("%d\n",N);
for(int i=0;i<N;i++){
            scanf("%*c %c %c %c%*c",&c0,&c1,&isD);
            edge0[i] = c0 -'A';
            edge1[i] = c1 -'A';
            isDes[i] = (isD =='x');
        }
        
        
        /*
        for(int i=0;i<N;i++){
            printf("%d, %d\n",edge0[i],edge1[1]);
        }*/
        
        cin >> Q;
        cout << cal(0,Q) << endl;
    }

//    while(1);
return0;
}

Summary: 简单的动态规划的题目。