POJ3278

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一维数组上的BFS 下一个状态点分别是 x-1， x + 1, 2 * x. x-1 之后需要判断是否小于0来剪枝

Source CodeProblem: 3278User: fistyMemory: 1512KTime: 47MSLanguage: G++Result: Accepted    Source Code    #include <cstdio>    #include <string>    #include <queue>    using namespace std;    int s,g;    #define MAX_N  200030    #define INF  0xffff    int dx[2] = {-1, 1};    int d[MAX_N];    void solve(){            queue<int> que;            fill(d, d + MAX_N, INF);            que.push(s);            d[s] = 0;            while(que.size()){                    int x = que.front(); que.pop();                    if(x == g) break;                    for(int i = 0;i < 3; i++){                            int nx;                            if(i == 2){                                    nx = x*2;                            }else{                                    nx = x + dx[i];                            }                            if(nx > MAX_N || nx < 0) continue;                            if(d[nx] == INF ){                                    que.push(nx);                                    d[nx] = d[x] + 1;                            }                    }                        }            printf("%d\n", d[g]);    }    int main(){            while(scanf("%d%d", &s, &g) != EOF){                    solve();            }            return 0;    }

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