CF 某套题 O ：Grid （简单BFS）

Posted 2020-10-29 ghzz

题意：

从左上角跳到右下角最少需要多少步，跳的规则为：可以向四个方向的任意一个方向跳当前格子中的步数，若跳不到右下角输出IMPOSSIBLE。

题解：

 BFS搜索，注意判断边界，标记。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <cstdlib>
#include <utility>
using namespace std;
#define is_lower(c) (c>=‘a‘ && c<=‘z‘)
#define is_upper(c) (c>=‘A‘ && c<=‘Z‘)
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c>=‘0‘ && c<=‘9‘)
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define IO ios::sync_with_stdio(0);\\
    cin.tie(0);    cout.tie(0);
#define For(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<int,int > pll;
typedef vector<int> vi;
const ll inf=0x3f3f3f3f;
const double EPS=1e-10;
const ll inf_ll=(ll)1e18;
const ll maxn=100005LL;
const ll mod=1000000007LL;
const int N = 500+5;
pair <int,int> p;
queue <pii> q;
int mp[N][N];
int xx[N][N];
bool vis[N][N];
int main() {
    int m,n;
    cin >> m >> n;
    For(i,1,m)
        For(j,1,n)
            scanf("%1d",&mp[i][j]);
    p.first = 1;p.second = 1;
    q.push(p);
    vis[1][1] = 1;
    int sum = 0;
    while(!q.empty()){
        pii x = q.front();
        q.pop();
        int step = mp[x.first][x.second];
        pii x1;
        if((x.first+step)>=1&&(x.first+step)<=m&&!vis[x.first+step][x.second]){
            x1.first = x.first+step,x1.second = x.second;
            q.push(x1);
            xx[x1.first][x1.second] = xx[x.first][x.second]+1;
            vis[x1.first][x1.second] = 1;
        }//向下
        if((x.first-step)>=1&&(x.first-step)<=m&&!vis[x.first-step][x.second]){
            x1.first = x.first-step,x1.second = x.second;
            q.push(x1);
            xx[x1.first][x1.second] = xx[x.first][x.second]+1;
            vis[x1.first][x1.second] = 1;
        }// up
        if((x.second+step)>=1&&(x.second+step)<=n&&!vis[x.first][x.second+step]){
            x1.first = x.first,x1.second = x.second+step;
            q.push(x1);
            xx[x1.first][x1.second] = xx[x.first][x.second]+1;
            vis[x1.first][x1.second] = 1;
        }
        if((x.second-step)>=1&&(x.second-step)<=n&&!vis[x.first][x.second-step]){
            x1.first = x.first,x1.second = x.second-step;
            q.push(x1);
            xx[x1.first][x1.second] = xx[x.first][x.second]+1;
            vis[x1.first][x1.second] = 1;
        }
        if(xx[m][n]!=0)
            break;
    }
    if(xx[m][n]!=0)
        cout << xx[m][n] <<endl;
    else
        cout <<"IMPOSSIBLE" << endl;
    return 0;
}