Codeforces 41d pawn simple DP

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Matrix constant K for N * m

Next is a matrix of N * M. Each position is represented by an integer of 0-9.

Q:

Go up from the last line to the first line so that the sum of % (k + 1) on the path is = 0

Each grid can only take one step or step.

The greatest path and

Position of the last row as the starting point

Bottom-to-top path

Ideas:

Simple DP

#include <cstdio>#include <algorithm>#include<iostream>#include<string.h>#include <math.h>#include<queue>#include<map>#include<vector>#include<set>using namespace std;#define N 105#define inf 10000000#define ll intint n,m,k;int dp[N][N][12];int px[N][N][12], py[N][N][12], sum[N][N][12];int mp[N][N];vector<pair<int,int> >ans;int main(){    int i, j, z;    while(~scanf("%d %d %d",&n,&m,&k)){            k++;        memset(sum, -1, sizeof sum);        memset(px, -1, sizeof px);        memset(py, -1, sizeof py);        memset(dp, 0, sizeof dp);        for(i = 1; i <= n; i++)            for(j = 1; j <= m; j++)            scanf("%1d",&mp[i][j]);        for(i = 1; i <= m; i++)        dp[n][i][mp[n][i] % k] = 1, sum[n][i][mp[n][i] % k] = mp[n][i];        for(i = n-1; i ; i--){            for(j = 1; j <= m; j++) {                int x = i+1, y = j-1;                if(y>=1)                {                    for(z = 0; z <= k; z++)                    if(dp[x][y][z] && sum[i][j][(z+mp[i][j])%k] < sum[x][y][z]+mp[i][j])                    {                        dp[i][j][(z+mp[i][j])%k] = 1;                        px[i][j][(z+mp[i][j])%k] = x;                        py[i][j][(z+mp[i][j])%k] = y;                        sum[i][j][(z+mp[i][j])%k] = sum[x][y][z]+mp[i][j];                    }                }                y = j+1;                if(y<=m)                {                    for(z = 0; z <= k; z++)                    if(dp[x][y][z] && sum[i][j][(z+mp[i][j])%k] < sum[x][y][z]+mp[i][j])                    {                        dp[i][j][(z+mp[i][j])%k] = 1;                        px[i][j][(z+mp[i][j])%k] = x;                        py[i][j][(z+mp[i][j])%k] = y;                        sum[i][j][(z+mp[i][j])%k] = sum[x][y][z]+mp[i][j];                    }                }            }        }        int posx = 1, posy = -1, mod = 0, anssum = -1;        for(i = 1; i <= m; i++)            if(dp[1][i][0] && anssum<sum[1][i][0])                posy = i, anssum = sum[1][i][0];        if(posy==-1){puts("-1");continue;}        ans.clear();        while(posy!=-1) {            ans.push_back(pair<int,int>(posx, posy));            int tx = px[posx][posy][mod];            int ty = py[posx][posy][mod];            mod = ((mod-mp[posx][posy])%k+k)%k;            posx = tx, posy = ty;        }        cout<<anssum<<endl;        int x = ans[ans.size()-1].first, y = ans[ans.size()-1].second;        cout<<y<<endl;        for(i = ans.size()-2; i >= 0; i--){            int nowx = ans[i].first, nowy = ans[i].second;            if(nowy>y)                printf("R");            else printf("L");            x = nowx, y = nowy;        }        puts("");    }    return 0;}