E - Palindromic Shortest Path【ABC394】

E - Palindromic Shortest Path

思路：

bfs，队列里存边即可。根据回文的特性，从长度为1和2的两种边开始找，往两边扩张。
要注意入队顺序，先入长度为1的边，完了再入长度为2的边，否则不能保证先找到的是最短的了。

代码：

//赛时写的史山
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
#define int long long
#define pb push_back
#define pii pair<int,int>
#define FU(i, a, b) for(int i = (a); i <= (b); ++ i)
#define FD(i, a, b) for(int i = (a); i >= (b); -- i)
const int MOD = 1e9+7;
const int INF = 1e9;
char c[105][105];
int ans[105][105];
bool vis[105][105];

signed main() {
    // cin.tie(0)->ios::sync_with_stdio(0);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin>>c[i][j];
            ans[i][j]= (i==j?0:INF);
        }
    }

    queue<pair<pii,int>> qu;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(c[i][j]=='-')continue;

            qu.push({{i,j},1});     // 奇数型
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(c[i][j]=='-')continue;
            for(int k=1;k<=n;k++){
                if(c[i][j]==c[j][k] && i!=k){
                    qu.push({{i,k},2});    //偶数型
                }
            }
        }
    }

    while(!qu.empty()){
        int i=qu.front().first.first,j=qu.front().first.second,k=qu.front().second;
        qu.pop();
        // cout<<i<<" "<<j<<" "<<k<<endl;
        ans[i][j]=min(ans[i][j],k);
        if(vis[i][j])continue;
        vis[i][j]=1;

        for(int l1=1;l1<=n;l1++){
            if(c[l1][i]=='-')continue;
            for(int l2=1;l2<=n;l2++){
                if(c[l1][i]== c[j][l2]){
                    qu.push({{l1,l2},k+2});
                }
            }
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(ans[i][j]!=INF)
            cout<<ans[i][j]<<" ";
            else cout<<"-1 ";
        }
        cout<<endl;
    }
    
    return 0;
}	

标答代码

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int inf = 1000000010;

int main() {
	int n;
	cin >> n;
	vector<vector<char>> c(n, vector<char>(n));
	rep(i, n) rep(j, n) cin >> c[i][j];
	vector<vector<int>> a(n, vector<int>(n, inf));
	queue<pair<int, int>> que;
	rep(i, n) {
		que.push({i, i});
		a[i][i] = 0;
	}
	rep(i, n) rep(j, n) {
		if (i == j or c[i][j] == '-') continue;
		que.push({i, j});
		a[i][j] = 1;
	}
	while (!que.empty()) {
		auto q = que.front(); que.pop();
		int i = q.first, j = q.second;
		rep(k, n) rep(l, n) {
			if (c[k][i] != '-' && c[j][l] != '-' && c[k][i] == c[j][l] && a[k][l] == inf) {
				a[k][l] = a[i][j] + 2;
				que.push({k, l});
			}
		}
	}
	rep(i, n) {
		rep(j, n) {
			cout << (a[i][j] == inf ? -1 : a[i][j]) << " \n"[j == n - 1];
		}
	}
}