CF 2B The Least round Way (DP) __acm_codeforces

Reprint please indicate the source, thank you http://blog.csdn.net/acm_cxlove?viewmode=contents    by---cxlove
topic: A n*n matrix, Each time you can only go down or to the right, from top left to right, a path of all the numbers multiplied, the end of the judgment at least several 0 http://codeforces.com/problemset/problem/2/B 
See the fork in the group recommended, and did, You deserve to have it. The end of the multiplication is 0, the description is 2*5, the final product can be expressed as 2^a  *  5^b  *  other, then the number of the last 0 is min (a,b)   Dp[i][j][0] A situation that represents the fewest number of record factor 2 dp[i][j][1] represents the least number of record factor 5 but there is a place where there is a trick that the number of matrices can be 0 product is 0.
So first 0 as 10 run a DP, if the result is 0, indicating that there is a path does not go through 0, and there is no 0 at the end if the result is greater than 1, you can choose this path through 0, then the answer is 1

#include <iostream> #include <cstdio> #include <map> #include <cstring> #include <cmath> #include <vector> #include <algorithm> #include <set> #include <string> #include <queue&gt  
; #define INF 1000000005 #define M #define N 10005 #define MAXN 300005 #define EPS 1e-8 #define ZERO (a) fabs (a) < EPS #define MIN (A,B) ((a) < (b)? ( A):(B) #define MAX (A,b) ((a) > (b)? ( A):(B) #define PB (a) push_back (a) #define MP (A,B) Make_pair (a,b) #define MEM (a,b) memset (A,b,sizeof (a)) #define L L Long Long #define MOD 1000000009 #define Lson step<<1 #define Rson step<<1|1 #define SQR (a) ((a) * (a)) #d  
Efine key_value ch[ch[root][1]][0] #define Test puts ("OK"); #define PI ACOs ( -1.0) #define LOWBIT (x) ((-(x)) & (x)) #define HASH1 1331 #define HASH2 10001 #pragma comment (linker, "/
stack:1024000000,1024000000 ") using namespace std;
DP[I][J][K][L] indicates arrival (I,J), l=0 means two,l=1 represents five int dp[1005][1005][2][2];int n,a[1005][1005];
pair<int,pair<int,int> > Pre[1005][1005][2];
int way[2][2]={0,1,1,0};
    int get (int num,int FAC) {int ret=0;
    if (!num) return 1;
        while (num&& (NUM%FAC) ==0) {ret++;
    NUM/=FAC;
return ret;
    int main () {//freopen ("Input.txt", "R", stdin);
        while (scanf ("%d", &n)!=eof) {int zx=-1,zy=-1;
                for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) {scanf ("%d", &a[i][j]);
            if (a[i][j]==0) zx=i,zy=j;
        } mem (dp,-1);
        for (int j=0;j<2;j++) for (int k=0;k<2;k++) dp[0][1][j][k]=dp[1][0][j][k]=0; for (int i=1;i<=n;i++) {for (int j=1;j<=n;j++) {int two=get (a[i][j],2), Five=get (a[i][j],5)
                ;
                    for (int r=0;r<2;r++) {int x=i-way[r][0],y=j-way[r][1]; for (int k=0;k<2;k++) {if (dp[x][y][k][0]==-1| |
                        DP[X][Y][K][1]==-1) continue;
                        int now_two=dp[x][y][k][0]+two,now_five=dp[x][y][k][1]+five;
                            if (dp[i][j][k][0]==-1) {dp[i][j][k][0]=now_two;
                            dp[i][j][k][1]=now_five;
                        PRE[I][J][K]=MP (K,MP (x,y)); else if (K&&now_two<dp[i][j][k][0]) {DP[I][J][K][0]=NOW_TW
                            O
                            dp[i][j][k][1]=now_five;
                        PRE[I][J][K]=MP (K,MP (x,y)); else if (!k&&now_five<dp[i][j][k][1]) {Dp[i][j][k][0]=now_
                            two;
                            dp[i][j][k][1]=now_five;
                        PRE[I][J][K]=MP (K,MP (x,y));
        ' {}}}} int ans=inf,k; for (int i=0;i<2;i+ +) {if (dp[n][n][i][0]==-1) continue;
            int tmp=min (dp[n][n][i][0],dp[n][n][i][1]);
                if (Tmp<ans) {ans=tmp;
            K=i;
            } if (ans>1&&zx!=-1) {printf ("1\n");
            for (int i=1;i<zx;i++) Putchar (' D ');
            for (int j=1;j<zy;j++) Putchar (' R ');
            for (int i=zx;i<n;i++) Putchar (' D ');
            for (int j=zy;j<n;j++) Putchar (' R ');
        Continue
        } string ret= "";
        int x=n,y=n; while (x!=1| |
            y!=1) {int xx=pre[x][y][k].second.first,yy=pre[x][y][k].second.second;
            if (xx==x) ret+= ' R ';
            else if (yy==y) ret+= ' D ';
        K=pre[x][y][k].first;x=xx;y=yy;
        } reverse (Ret.begin (), Ret.end ());
    cout<<ans<<endl<<ret<<endl;
return 0; }