CF 2B The least round way
2016-10-25 20:07
405 查看
题目大意:一个n∗n的矩阵M,每个位置有一个值Mi,j,从左上角出发到右下角,只能向右或向下走,并将途经的数字相乘,求一条路径使得乘积的结尾零最少。
题解:将每一个数字分解,看有多少个2和多少个5。那么问题就变成了求从左上到右下经过最少的2或最少的5的路径。dp即可。注意,当有0存在时,如果结尾0大于1,那么一定会走过0的一条路径。
题解:将每一个数字分解,看有多少个2和多少个5。那么问题就变成了求从左上到右下经过最少的2或最少的5的路径。dp即可。注意,当有0存在时,如果结尾0大于1,那么一定会走过0的一条路径。
#include <bits/stdc++.h> using namespace std; int a[1005][1005],b[1005][1005]; int dp[1005][1005][2][2]; int st[1005][1005][2]; void print(int x,int y,int op){ if(x == 0 && y == 0){ return ; } if(st[x][y][op]){ print(x-1,y,op); printf("D"); } else{ print(x,y-1,op); printf("R"); } return ; } int main(){ int n; scanf("%d",&n); memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(st,0,sizeof(st)); int fi,fj; fi = fj = -1; for(int i = 0;i < n;i++){ for(int j = 0;j < n;j++){ int x; scanf("%d",&x); if(!x){ fi = i; fj = j; } while(x&&x%2==0) { x>>=1; a[i][j]++; } while(x&&x%5==0){ x/=5; b[i][j]++; } } } for(int i = 0;i < 2;i++){ dp[0][0][i][0] = a[0][0]; dp[0][0][i][1] = b[0][0]; } for(int i = 1;i < n;i++){ for(int j = 0;j < 2;j++){ dp[i][0][j][0] = dp[i-1][0][j][0]+a[i][0]; dp[i][0][j][1] = dp[i-1][0][j][1]+b[i][0]; st[i][0][j] = 1; dp[0][i][j][0] = dp[0][i-1][j][0]+a[0][i]; dp[0][i][j][1] = dp[0][i-1][j][1]+b[0][i]; } } for(int i = 1;i < n;i++){ for(int j = 1;j < n;j++){ if(dp[i][j-1][0][0] == dp[i-1][j][0][0]){ dp[i][j][0][0] = dp[i][j-1][0][0]+a[i][j]; if(dp[i-1][j][0][1]<dp[i][j-1][0][1]) st[i][j][0] = 1; dp[i][j][0][1] = min(dp[i-1][j][0][1],dp[i][j-1][0][1])+b[i][j]; } else if(dp[i][j-1][0][0] < dp[i-1][j][0][0]){ dp[i][j][0][0] = dp[i][j-1][0][0]+a[i][j]; dp[i][j][0][1] = dp[i][j-1][0][1]+b[i][j]; } else{ dp[i][j][0][0] = dp[i-1][j][0][0]+a[i][j]; dp[i][j][0][1] = dp[i-1][j][0][1]+b[i][j]; st[i][j][0] = 1; } if(dp[i][j-1][1][1] == dp[i-1][j][1][1]){ dp[i][j][1][1] = dp[i][j-1][1][1]+b[i][j]; if(dp[i-1][j][1][0]<dp[i][j-1][1][0]) st[i][j][1] = 1; dp[i][j][1][0] = min(dp[i-1][j][1][0],dp[i][j-1][1][0])+a[i][j]; } else if(dp[i][j-1][1][1] < dp[i-1][j][1][1]){ dp[i][j][1][1] = dp[i][j-1][1][1]+b[i][j]; dp[i][j][1][0] = dp[i][j-1][1][0]+a[i][j]; } else{ dp[i][j][1][1] = dp[i-1][j][1][1]+b[i][j]; dp[i][j][1][0] = dp[i-1][j][1][0]+a[i][j]; st[i][j][1] = 1; } } } /*for(int i = 0;i < n;i++){ for(int j= 0;j < n;j++){ printf("%d/%d ",a[i][j],b[i][j]); } printf("\n"); } cout<<endl;*/ /*for(int k = 0;k < 2;k++){ for(int i = 0;i < n;i++){ for(int j = 0;j < n;j++){ printf("%d/%d ",dp[i][j][k][0],dp[i][j][k][1]); } printf("\n"); } printf("\n"); }*/ if(fi!= -1){ if(min(min(dp[n-1][n-1][0][0],dp[n-1][n-1][0][1]),min(dp[n-1][n-1][1][0],dp[n-1][n-1][1][1]))>1){ printf("1\n"); for(int i = 0;i < fi;i++) printf("D"); for(int j = 0;j < fj;j++) printf("R"); for(int i = fi;i < n-1;i++) printf("D"); for(int i = fj;i < n-1;i++) printf("R"); return 0; } } if(min(dp[n-1][n-1][0][0],dp[n-1][n-1][0][1])<min(dp[n-1][n-1][1][0],dp[n-1][n-1][1][1])){ printf("%d\n",min(dp[n-1][n-1][0][0],dp[n-1][n-1][0][1])); print(n-1,n-1,0); } else{ printf("%d\n",min(dp[n-1][n-1][1][0],dp[n-1][n-1][1][1])); print(n-1,n-1,1); } return 0; }
相关文章推荐
- CF 2B.The least round way
- CF_2B_TheLeastRoundWay
- CF 2B The least round way(DP)
- CF 2B The least round way
- Codeforces 2B The least round way
- codefroces 2B The least round way
- codeforces 2B.The least round way(数学&dp)
- Codefoeces 2B. The least round way
- [CF2B]The least round way
- 2B The least round way
- Codeforces 2B The least round way
- codeforces 2B The least round way
- CodeForces 2B The least round way
- [codeforces] 2B - The least round way
- 最小较小codeforces 2B The least round way
- CodeForces 2B The least round way(dp+数学)
- CF 2 B. The least round way
- 2B - The least round way
- codeforces 2B The least round way
- coderforces 2B the least round way