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。

 

  1 #include <cstring>

  2 #include <cstdio>

  3 #include <string>

  4 #include <iostream>

  5 #include <algorithm>

  6 #include <vector>

  7 #include <queue>

  8 using namespace std;

  9 #define INF 0xfffffff

 10 vector<string>::iterator it;

 11 int p[101][101];

 12 int a[4] = {0,0,1,-1};

 13 int b[4] = {1,-1,0,0};

 14 struct node

 15 {

 16     int u,v,w,next;

 17 }edge[1000001];

 18 int first[2001];

 19 int in[2001],d[2001];

 20 int tot;

 21 void CL()

 22 {

 23     memset(first,-1,sizeof(first));

 24     tot = 0;

 25 }

 26 void add(int u,int v,int w)

 27 {

 28     edge[tot].u = u;

 29     edge[tot].v = v;

 30     edge[tot].w = w;

 31     edge[tot].next = first[u];

 32     first[u] = tot ++;

 33 }

 34 int spfa(int x,int n,int key)

 35 {

 36     int i,u,v;

 37     queue<int> que;

 38     for(i = 1;i <= n;i ++)

 39     {

 40         in[i] = 0;

 41         d[i] = INF;

 42     }

 43     in[x] = 0;

 44     d[x] = key;

 45     que.push(x);

 46     while(!que.empty())

 47     {

 48         u = que.front();

 49         que.pop();

 50         in[u] = 0;

 51         for(i = first[u];i != -1;i = edge[i].next)

 52         {

 53             v = edge[i].v;

 54             if(d[v] > d[u] + edge[i].w)

 55             {

 56                 d[v] = d[u] + edge[i].w;

 57                 if(!in[v])

 58                 {

 59                     in[v] = 1;

 60                     que.push(v);

 61                 }

 62             }

 63         }

 64     }

 65     int ans = 0;

 66     for(i = 1;i <= n;i ++)

 67     {

 68         if(i != x)

 69         ans = max(ans,d[i]);

 70     }

 71     return ans;

 72 }

 73 class GameOnABoard

 74 {

 75 public:

 76     int optimalChoice(vector <string> cost)

 77     {

 78         int i,j,k,n,len,ans;

 79         n = 1;

 80         CL();

 81         for(it = cost.begin();it != cost.end();it ++)

 82         {

 83             len = 0;

 84             for(j = 1;j <= (*it).size();j ++)

 85             {

 86                 p[n][j] = (*it)[j-1] - '0';

 87                 len ++;

 88             }

 89             n ++;

 90         }

 91         n --;

 92         for(i = 1;i <= n;i ++)

 93         {

 94             for(j = 1;j <= len;j ++)

 95             {

 96                 for(k = 0;k < 4;k ++)

 97                 {

 98                     if(i+a[k] >= 1&&i + a[k] <= n&&j + b[k] >= 1&&j + b[k] <= len)

 99                     {

100                         add((i-1)*len+j,(i+a[k]-1)*len+j+b[k],p[i+a[k]][j+b[k]]);

101                     }

102                 }

103             }

104         }

105         ans = 1000000000;

106         for(i = 1;i <= n;i ++)

107         {

108             for(j = 1;j <= len;j ++)

109             ans = min(ans,spfa((i-1)*len+j,n*len,p[i][j]));

110         }

111         return ans;

112     }

113 };