TC SRM 583 DIV 2

做了俩，rating涨了80。第二个题是关于身份证的模拟题，写的时间比较长，但是我认真检查了。。。

第三个题是最短路，今天写了写，写的很繁琐，写的很多错。

 #include <cstring>
 #include <cstdio>
 #include <string>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 using namespace std;
 #define INF 0xfffffff
 vector<string>::iterator it;
 int p[][];
 int a[] = {,,,-};
 int b[] = {,-,,};
 struct node
 {
     int u,v,w,next;
 }edge[];
 int first[];
 int in[],d[];
 int tot;
 void CL()
 {
     memset(first,-,sizeof(first));
     tot = ;
 }
 void add(int u,int v,int w)
 {
     edge[tot].u = u;
     edge[tot].v = v;
     edge[tot].w = w;
     edge[tot].next = first[u];
     first[u] = tot ++;
 }
 int spfa(int x,int n,int key)
 {
     int i,u,v;
     queue<int> que;
     for(i = ;i <= n;i ++)
     {
         in[i] = ;
         d[i] = INF;
     }
     in[x] = ;
     d[x] = key;
     que.push(x);
     while(!que.empty())
     {
         u = que.front();
         que.pop();
         in[u] = ;
         for(i = first[u];i != -;i = edge[i].next)
         {
             v = edge[i].v;
             if(d[v] > d[u] + edge[i].w)
             {
                 d[v] = d[u] + edge[i].w;
                 if(!in[v])
                 {
                     in[v] = ;
                     que.push(v);
                 }
             }
         }
     }
     int ans = ;
     for(i = ;i <= n;i ++)
     {
         if(i != x)
         ans = max(ans,d[i]);
     }
     return ans;
 }
 class GameOnABoard
 {
 public:
     int optimalChoice(vector <string> cost)
     {
         int i,j,k,n,len,ans;
         n = ;
         CL();
         for(it = cost.begin();it != cost.end();it ++)
         {
             len = ;
             for(j = ;j <= (*it).size();j ++)
             {
                 p[n][j] = (*it)[j-] - '';
                 len ++;
             }
             n ++;
         }
         n --;
         for(i = ;i <= n;i ++)
         {
             for(j = ;j <= len;j ++)
             {
                 for(k = ;k < ;k ++)
                 {
                     if(i+a[k] >= &&i + a[k] <= n&&j + b[k] >= &&j + b[k] <= len)
                     {
                         add((i-)*len+j,(i+a[k]-)*len+j+b[k],p[i+a[k]][j+b[k]]);
                     }
                 }
             }
         }
         ans = ;
         for(i = ;i <= n;i ++)
         {
             for(j = ;j <= len;j ++)
             ans = min(ans,spfa((i-)*len+j,n*len,p[i][j]));
         }
         return ans;
     }
 };