codeforces 590C(bfs)

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本体题意不再描述。

把三个国家缩成三个点，那么把这三个点连起来，只有两种方法，可以尝试画一下，然后思路就很明显了。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>using namespace std;typedef long long ll;#define rep1(i,x,y) for(int i=x;i<=y;i++)#define rep(i,n) for(int i=0;i<(int)n;i++)#define clr(a,x) memset((a),(x),sizeof(a))const int inf = 1e7;const int N = 1010;int n,m;struct node{   int u,x,y;   node(){}   node(int u=0,int x=0,int y=0):u(u),x(x),y(y){}};char ma[N][N];int dis[3][N][N];queue<node> Q;const int dx[]={1,0,-1,0};const int dy[]={0,1,0,-1};int judge(int i,int j){return i>=1&&i<=n&&j>=1&&j<=m;}int goal;void relax(int u , int nx,int ny,int add){    if(dis[u][nx][ny]==inf){        dis[u][nx][ny] = add; Q.push(node(u,nx,ny));    }}int vis[N][N]={0};int dist[3][3];int bfs(int p){   while(!Q.empty()) Q.pop();   rep1(i,1,n) rep1(j,1,m) {       dis[p][i][j]=inf;       if(ma[i][j]-'1' == p) {            Q.push(node(p,i,j));            dis[p][i][j] = 0;       }   }   while(!Q.empty()){       node u =Q.front();  Q.pop();       rep(i,4){          int nx=u.x+dx[i],ny=u.y+dy[i];          if(judge(nx,ny) && ma[nx][ny]=='.'){             relax(p,nx,ny,dis[p][u.x][u.y] + 1);          }          if(judge(nx,ny) && isdigit(ma[nx][ny])){                 int to = ma[nx][ny]-'1';                 if(dist[p][to] == inf)                 dist[p][to] = dis[p][u.x][u.y];          }       }   }}int main(){   rep(i,3) rep(j,3) dist[i][j] = inf;   scanf("%d %d",&n,&m);   rep1(i,1,n)  scanf("%s",ma[i]+1);   int ans = inf;   rep(i,3) bfs(i);   ans = dist[0][1]+dist[0][2]+dist[1][2]-max(dist[0][1],max(dist[0][2],dist[1][2]));   rep1(i,1,n) rep1(j,1,m)if(ma[i][j] == '.'){       ans = min(ans , dis[0][i][j]+dis[1][i][j]+dis[2][i][j]-2);   }    printf("%d\n",ans>=inf ? -1 : ans);   return 0;}

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