ICPC Pacific Northwest Regional Contest 2016 C. Buggy Robot

Buggy Robot　　

思路：dp[inx][x][y],表示用了前inx个指令后的最小费用。

对于一个指令，我们可以选择不走或者走，其他的我们可以添加四个方向的指令与使用过指令后的dp来比较。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

const int N = 60;
const int INF = 1e9;
int mv_x[] = {1, -1, 0, 0};
int mv_y[] = {0, 0, -1, 1};
char mp[N][N];
int dp[N][N][N];
char str[N];
int n, m;

struct node
{
    int inx, x, y;
}R, E;

bool inline check(int x, int y)
{
    return x >= 0 && x < n && y >= 0 && y < m;
}

void bfs()
{
    for(int i = 0; i < N; ++i)
        for(int j = 0; j < N; ++j)
            for(int k = 0; k < N; ++k)
                dp[i][j][k] = INF;

    int len = strlen(str);
    queue<node > que; 

    dp[0][R.x][R.y] = 0;
    que.push({0, R.x, R.y});

    while(!que.empty()){
        node now = que.front();
        que.pop();

        for(int p = 0; p < 4; ++p){
            int dx = now.x + mv_x[p];
            int dy = now.y + mv_y[p];

            if(check(dx, dy) && mp[dx][dy] != '#'){
                if(dp[now.inx][now.x][now.y] + 1 < dp[now.inx][dx][dy]){
                    dp[now.inx][dx][dy] = dp[now.inx][now.x][now.y] + 1;
                    que.push({now.inx, dx, dy});
                }
            }

            if(now.inx < len){
                //不走
                if(dp[now.inx][now.x][now.y] + 1 < dp[now.inx + 1][now.x][now.y]){
                    dp[now.inx + 1][now.x][now.y] = dp[now.inx][now.x][now.y] + 1;
                    que.push({now.inx + 1, now.x, now.y}); 
                }

                //走
                dx = now.x; 
                dy = now.y;
                if(str[now.inx] == 'L') dy--;
                if(str[now.inx] == 'R') dy++;
                if(str[now.inx] == 'U') dx--;
                if(str[now.inx] == 'D') dx++;
                if(!check(dx,dy) || mp[dx][dy] == '#') dx = now.x, dy = now.y;
                if(dp[now.inx][now.x][now.y] < dp[now.inx + 1][dx][dy]){
                    dp[now.inx + 1][dx][dy] = dp[now.inx][now.x][now.y];
                    que.push({now.inx + 1, dx, dy}); 
                }                
            }
        }
    }
}

void solve()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; ++i) scanf("%s", mp[i]);
    scanf("%s", str);
    //cout << "len = " << (strlen(str)) << endl;
    
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            if(mp[i][j] == 'R'){
                R.x = i; R.y = j;
            }
            if(mp[i][j] == 'E'){
                E.x = i; E.y = j;
            }
        }
    }

    bfs();

    int ans = INF;
    for(int i = 0; i < N; ++i){
        ans = min(ans, dp[i][E.x][E.y]);
    }
    //printf("ans = %d
", ans);
    printf("%d
", ans);
}


int main()
{

    solve();

    return 0;
}