P6370. [COCI2006-2007#6] KAMEN

60 分

#include <bits/stdc++.h>
using namespace std;
const int MAXR = 30000;
const int MAXC = 30;
int r, c, n;
char g[MAXR + 5][MAXC + 5];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> r >> c;
    for (int i = 1; i <= r; i++)
        for (int j = 1; j <= c; j++)
            cin >> g[i][j];
    cin >> n;
    while (n--)
    {
        int x, y;
        cin >> y;
        x = 1;
        if (g[x][y] == 'X' || g[x][y] == 'O')
            continue;
        while (x != r)
        {
            if (g[x + 1][y] == '.')
                x++;
            else if (g[x + 1][y] == 'X')
                break;
            else if (y != 1 && g[x][y - 1] == '.' && g[x + 1][y - 1] == '.')
                x++, y--;
            else if (y != c && g[x][y + 1] == '.' && g[x + 1][y + 1] == '.')
                x++, y++;
            else
                break;
        }
        g[x][y] = 'O';
    }
    for (int i = 1; i <= r; i++)
    {
        for (int j = 1; j <= c; j++)
            cout << g[i][j];
        cout << "\n";
    }
    return 0;
}

100 分

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int r, c, m;
//存图
char g[30005][35];
//(1,i)为起点的路径，第j个点为(j,path[i][j])
int path[35][30005];
//(1,i)为起点的路径，最后一个点在第lastRow[i]行
int lastRow[35];
//维护路径
void run(int ith, int nowX, int nowY)
{
    path[ith][nowX] = nowY;
    while (1)
    {
        if (g[nowX + 1][nowY] == 'X')
            break;
        if (g[nowX + 1][nowY] == '.')
            nowX++;
        else if (g[nowX][nowY - 1] == '.' && g[nowX + 1][nowY - 1] == '.')
            nowX++, nowY--;
        else if (g[nowX][nowY + 1] == '.' && g[nowX + 1][nowY + 1] == '.')
            nowX++, nowY++;
        else
            break;
        path[ith][nowX] = nowY;
    }
    lastRow[ith] = nowX;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> r >> c;
    memset(g, 'X', sizeof(g));
    for (int i = 1; i <= r; i++)
        for (int j = 1; j <= c; j++)
            cin >> g[i][j];
    for (int i = 1; i <= c; i++)
        run(i, 1, i);
    cin >> m;
    while (m--)
    {
        int nowC;
        cin >> nowC;
        int toX, toY;
        toX = lastRow[nowC];
        toY = path[nowC][toX];
        g[toX][toY] = 'O';
        for (int i = 1; i <= c; i++)
            if (lastRow[i] == toX && path[i][toX] == toY)
                run(i, toX - 1, path[i][toX - 1]);
    }
    for (int i = 1; i <= r; i++)
    {
        for (int j = 1; j <= c; j++)
            cout << g[i][j];
        cout << endl;
    }
    return 0;
}