Buggy Robot （二维压一维最短路

来源：互联网发布：人工智能取代人类辩论编辑：程序博客网时间：2024/06/08 07:55
题意：可以任意删除或添加指令。最后输出修改了多少次指令，如果要撞墙就不走。
#include<bits/stdc++.h>using namespace std;#define pi acos(-1)#define endl '\n'#define rand() srand(time(0));#define me(x) memset(x,0,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0);typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,-1,-1,1,1};const int dy[]={0,1,0,-1,1,-1,1,-1};const int maxn=1e4+5;const int maxx=1e6+100;const double EPS=1e-7;const int MOD=10000007;typedef pair<int, int>P;#define mod(x) ((x)%MOD);template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W whilevector<pair<int,int> >E[maxx];int n,m;int d[maxx],inq[maxx];char mapp[55][55],op[55];int check(int x,int y){    if(x>=1&&x<=n&&y>=1&&y<=m&&mapp[x][y]!='#')        return 1;    return 0;}int main(){    cin>>n>>m;    FOR(1,n,i)        cin>>mapp[i]+1;    cin>>op+1;    for(int i=0;i<maxx;i++)    {        E[i].clear();        inq[i]=0;        d[i]=INF;    }    int st,ed,len=strlen(op+1);    FOR(1,n,i)        FOR(1,m,j)    {        if (mapp[i][j] == 'R')  st=(i-1)*m+j;        if (mapp[i][j] == 'E')  ed=(i-1)*m+j;        if (mapp[i][j] == '#') continue;        FOR(1,len,k)        {            FOR(0,3,p)            {                int x=i+dx[p];                int y=j+dy[p];                if(check(x,y))                {                    int u=((i-1)*m+j)*(len+1)+k;                    int v=((x-1)*m+y)*(len+1)+k;                    E[u].push_back(make_pair(v,1));                   // cout<<u<<"  "<<v<<endl;                }            }            int x = i, y = j;            if (op[k] == 'R') y++;            else if (op[k] == 'L') y--;            else if (op[k] == 'U') x--;            else x++;            if(check(x,y))            {                int u=((i-1)*m+j)*(len+1)+k;                int v=((x-1)*m+y)*(len+1)+k+1;                E[u].push_back(make_pair(v,0));            }            else            {                int u=((i-1)*m+j)*(len+1)+k;                E[u].push_back(make_pair(u+1,0));            }            //int u=((i-1)*m+j)*(len+1)+k;           // E[u].push_back(make_pair(u+1,1));            FOR(0,3,p)            {                int x=i+dx[p];                int y=j+dy[p];                if(check(x,y))                {                    int u=((i-1)*m+j)*(len+1)+len+1;                    int v=((x-1)*m+y)*(len+1)+len+1;                    E[u].push_back(make_pair(v,1));                }            }        }    }    queue<int >Q;    st=st*(len+1)+1;    Q.push(st),d[st]=0,inq[st]=1;    while(!Q.empty())    {        int now=Q.front();        Q.pop();        inq[now]=0;        for(int i=0;i<E[now].size();i++)        {            int v=E[now][i].first;            if(d[v]>d[now]+E[now][i].second)            {                d[v]=d[now]+E[now][i].second;                if(inq[v]==1) continue;                inq[v]=1;                Q.push(v);            }        }    }    int ans=INF;    FOR(1,len+1,i)    {        int cur=ed*(len+1)+i;        ans=min(ans,d[cur]);    }    cout<<ans<<endl;}
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