【洛谷】P1002 过河卒

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P1002 [NOIP2002 普及组] 过河卒

过河卒

思路

dp/动态规划
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A 0 0 0 0 0 0
0 0 X 0 X 0 0
0 X 0 0 0 X 0
0 0 0 M 0 0 0
0 X 0 0 0 X 0
0 0 X 0 X 0 0
0 0 0 0 0 0 B
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1 1 1 1 1 1 1
1 0 X 1 X 1 2
1 X 0 1 1 X 2
1 1 1 M 1 1 3
1 X 1 1 0 X 3
1 1 X 1 X 0 3
1 2 2 3 3 3 6

map[i][j] = map[i - 1][j] + map[i][j - 1];

dfs
  1. 得分 60 Unaccepted
  2. #3 TLE #4 TLE
  3. P1002_3.in 20 20 4 0
  4. P1002_3.out 56477364570

代码

dp
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#include <iostream>
#include <cstdio>
using namespace std;
int n, m, x, y;
long long map[42][42] = {{1}};
bool horse[42][42];
int hx[9] = {0, 1, 1, 2, 2, -1, -1, -2, -2},
    hy[9] = {0, 2, -2, 1, -1, 2, -2, 1, -1};
int main()
{
    cin >> n >> m >> x >> y;
    for (int i = 0; i <= 8; i++)
    {
        int xx = x + hx[i], yy = y + hy[i];
        if (xx > n || xx < 0 || yy > m || yy < 0)
            continue;
        horse[xx][yy] = 1;
    }
    int l = 0, k = 0;
    while (!horse[0][++k] && k <= m)
        map[0][k] = 1;
    while (!horse[++l][0] && l <= n)
        map[l][0] = 1;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (!horse[i][j])
                map[i][j] = map[i - 1][j] + map[i][j - 1];
    cout << map[n][m] << endl;
    return 0;
}
dfs+偏移量技巧 60分
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#include<bits/stdc++.h>
using namespace std;

const int N=21;
int dx[]={2,1,-1,-2,-2,-1,1,2,}, dy[]={1,2,2,1,-1,-2,-2,-1};
int ddx[]={0,1},ddy[]={1,0};
int p[N][N];
int n,m;
int res=0;

void init()
{
    for(int i=0;i<N;i++)
    {
        memset(p[i],0,sizeof(p[i]));
    }	
} 
void horse(int x, int y)
{
    p[x][y]=1;
    for(int i=0;i<8;i++)
    {
        int a=x+dx[i],b=y+dy[i];
        if(a >= 0 && a <= n && b >= 0 && b <= m)
        p[a][b]=1;
    }
}
void dfs(int x,int y)
{
    for(int i=0;i<2;i++)
    {
        int a=x+ddx[i],b=y+ddy[i];
        if(a==n and b==m)
        res++;
        if(a >= 0 && a <= n && b >= 0 && b <= m &&p[a][b]!=1)
            dfs(a,b);	
    }
}
int main()
{
    int x,y;
    cin>>n>>m>>x>>y;
    init();
    horse(x,y);
    dfs(0,0);
    cout<<res<<endl;
    return 0;
}