Codeforces 41D Pawn 简单dp

题目链接：点击打开链接

给定n*m 的矩阵 常数k

以下一个n*m的矩阵，每一个位置由 0-9的一个整数表示

问：

从最后一行開始向上走到第一行使得路径上的和 % (k+1) == 0

每一个格子仅仅能向↖或↗走一步

求：最大的路径和

最后一行的哪个位置作为起点

从下到上的路径

思路：

简单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 int
int 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;
}