poj3026

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题解

  • 这题很有意思,不用管A,S,当成一样的就好,跑一遍bfs跑出所有A的相关信息,然后套模板,然而我疯狂WA,原因是输入有坑,巨坑,
  • 详情看代码
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=107;
const int inf=0x3f3f3f3f;
typedef pair<int,int> s;
int T,n,m,dis[N][N],d[N][N],cnt,t[N][N],vis[N],ds[N];
int dx[5]={0,-1,1,0,0};
int dy[5]={0,0,0,-1,1};
char g[107][107];
queue<s> q;
void bfs(int a,int b){
    while(!q.empty())q.pop();
    memset(t,-1,sizeof(t));
    q.push(s(a,b));
    t[a][b]=0;
    while(!q.empty()){
        int x=q.front().first;
        int y=q.front().second;
        if(d[x][y])dis[d[a][b]][d[x][y]]=t[x][y];
        q.pop();
        rep(i,1,4){
            int nx=x+dx[i];
            int ny=y+dy[i];
            if(nx<1||nx>n||ny<1||ny>m||g[nx][ny]=='#'||t[nx][ny]!=-1)continue;
            t[nx][ny]=t[x][y]+1;
            q.push(s(nx,ny));
        }
    }
}

int prim(int n){
    int ans=0;
    memset(ds,0,sizeof(ds));
    memset(vis,0,sizeof(vis));
    vis[1]=1;
    rep(i,1,n)ds[i]=dis[1][i];
    rep(i,2,n){
        int minm=inf,p;
        rep(j,1,n)if(!vis[j]&&ds[j]<minm)minm=ds[j],p=j;
        if(minm==inf)return -1;
        ans+=minm;
        vis[p]=1;
        rep(j,1,n)if(!vis[j]&&ds[j]>dis[p][j])ds[j]=dis[p][j];
    }
    return ans;
}


int main(){
    scanf("%d",&T);
    while(T--){
        char c;
        cnt=0;
        scanf("%d%d",&m,&n);
        memset(d,0,sizeof(d));
        do{                         //坑点
            c=getchar();
        }while(c!='\n');
        rep(i,1,n){
            gets(g[i]+1);
            rep(j,1,m)if(g[i][j]=='S'||g[i][j]=='A')d[i][j]=++cnt;
        }
        rep(i,1,n)rep(j,1,m)if(d[i][j])bfs(i,j);
        printf("%d\n",prim(cnt));
    }
    return 0;
}