CodeForces 590C Three States BFS
题解:
以3个大陆为起点,都dfs一遍,求出该大陆到其他点的最小距离是多少, 然后枚举每个点作为3个大陆的路径交点。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define ep emplace_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; const int N = 1e3 + 100; int n, m; LL dis[N][N][3]; char s[N][N]; int dx[] = {1, -1, 0, 0}; int dy[] = {0, 0, -1, 1}; deque<pll> q; void bfs(int k){ for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) dis[i][j][k] = inf; char kk = k + '0' + 1; for(int i = 1; i <= n; ++i){ for(int j = 1; j <= m; ++j){ if(s[i][j] == kk){ dis[i][j][k] = 0; q.push_back(pll(i,j)); } } } while(!q.empty()){ pll t = q.front(); q.pop_front(); for(int _ = 0; _ < 4; ++_){ int nx = t.fi + dx[_]; int ny = t.se + dy[_]; if(nx < 1 || nx > n || ny < 1 || ny > m || s[nx][ny] == '#') continue; int dd = dis[t.fi][t.se][k] + (s[nx][ny] == '.'); if(dd < dis[nx][ny][k]){ dis[nx][ny][k] = dd; if(s[nx][ny] == '.') q.push_back(pll(nx,ny)); else q.push_front(pll(nx, ny)); } } } } int main(){ scanf("%d%d", &n, &m); for(int i = 1; i <= n; ++i) scanf("%s", s[i]+1); for(int i = 0; i < 3; ++i) bfs(i); LL ans = inf; for(int i = 1; i <= n; ++i){ for(int j = 1; j <= m; ++j){ int f = 0; if(s[i][j] == '.') f = 2; ans = min(ans, dis[i][j][0] + dis[i][j][1] + dis[i][j][2] - f); } } if(ans == inf) ans = -1; cout << ans << endl; return 0; }