Instantaneous Transference(强连通分量及其缩点)

http://poj.org/problem?id=3592

题意：给出一个n*m的矩阵，左上角代表起始点，每个格子都有一定价值的金矿，其中‘#’代表岩石不可达，‘*’代表时空门可以到达指定格子，求出可以获得的最大价值。

思路：时空门的存在可能会使得图中出现环，所以先对强连通分量进行缩点，然后对于缩点后的连通分量建立有向无环图，spfa求到起始点的最长路。

 #include <stdio.h>
 #include <iostream>
 #include <string.h>
 #include <queue>
 #include <stack>
 using namespace std;
 const int N=;
 const int INF=-<<;
 struct node
 {
     int u,v,w;
     int next;
 } edge1[N*],edge2[N*];//edge1[]存的是原始图的边，edge2[]存的是缩点后的边
 int low[N],dfn[N],sccno[N];//sccno[i]表示点i所属于的连通分量的编号
 int dis[N],head1[N],head2[N],val[N];//val[]存储每个连通分量的价值
 int n,m,scc_cnt,cnt,dfs_clock;
 bool vis[N];
 stack<int>S;

 void init()
 {
     cnt = ;
     scc_cnt = ;
     dfs_clock = ;
     memset(val,,sizeof(val));
     memset(low,,sizeof(low));
     memset(dfn,,sizeof(dfn));
     memset(sccno,,sizeof(sccno));
     memset(head1,-,sizeof(head1));
     memset(head2,-,sizeof(head2));

 }
 void add(int u,int v,int w,node *edge,int *head)
 {
     edge[cnt].u = u;
     edge[cnt].v = v;
     edge[cnt].w = w;
     edge[cnt].next = head[u];
     head[u] = cnt++;
 }
 void Tarjan(int u)//求强连通分量
 {
     low[u]=dfn[u]=++dfs_clock;
     S.push(u);
     for (int i = head1[u]; i!=-; i=edge1[i].next)
     {
         int v = edge1[i].v;
         if (!dfn[v])
         {
             Tarjan(v);
             low[u]=min(low[u],low[v]);
         }
         else if (!sccno[v])
         {
             low[u]=min(low[u],dfn[v]);
         }
     }
     if (low[u]==dfn[u])
     {
         ++scc_cnt;
         while()
         {
             int v = S.top();
             S.pop();
             sccno[v] = scc_cnt;
             if (v==u)
                 break;
         }
     }
 }
 void spfa(int s)//求缩点后的最长路
 {
     for (int i = ; i <= n*m; i++)
     {
         dis[i]=INF;
         vis[i]=false;
     }
     dis[s] = ;
     queue<int>q;
     q.push(s);
     vis[s] = true;
     while(!q.empty())
     {
         int u = q.front();
         q.pop();
         vis[u]=false;
         for (int i = head2[u]; i!=-; i=edge2[i].next)
         {
             int v = edge2[i].v;
             int w = edge2[i].w;
             if (dis[v] < dis[u]+w)
             {
                 dis[v] = dis[u]+w;
                 if (!vis[v])
                 {
                     q.push(v);
                     vis[v]=true;
                 }
             }

         }
     }
 }
 int main()
 {
     int t;
     char map[][];
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         init();
         for(int i = ; i < n; i++)
         {
             scanf("%s",map[i]);
         }
         for (int i = ; i < n; i++)
         {
             for (int j = ; j < m; j++)
             {
                 if (map[i][j]!='#')
                 {
                     if(i) add((i-)*m+j,i*m+j,,edge1,head1);
                     if(j) add(i*m+j-,i*m+j,,edge1,head1);
                     if (map[i][j]=='*')
                     {
                         int x,y;
                         scanf("%d%d",&x,&y);
                         if(map[x][y]!='#')
                             add(i*m+j,x*m+y,,edge1,head1);
                     }
                 }
             }
         }
         for (int i = ; i < n; i++)
         {
             if (!dfn[i])
                 Tarjan(i);
         }
         for (int i = ; i < n; i++)
         {
             for (int j = ; j < m; j++)
             {
                 if(map[i][j]!='*'&&map[i][j]!='#')
                 {
                     int num = sccno[i*m+j];
                     val[num]+=map[i][j]-'';//记录每个连通分量的价值
                 }
             }
         }
         for (int u = ; u < n*m; u++)
         {
             for (int j = head1[u]; j!=-; j=edge1[j].next)
             {
                 int v = edge1[j].v;
                 if(sccno[u]!=sccno[v])
                 {
                     add(sccno[u],sccno[v],val[sccno[v]],edge2,head2);
                 }//对缩点后的连通分量建立有向图
             }
         }
         spfa(sccno[]);//求出每个点到起始点的价值
         int max=-;
         for (int i = ; i <= scc_cnt; i++)
         {
             if (max < dis[i])
                 max = dis[i];//求出到原点的最大价值
         }
         int ans = max+val[sccno[]];//总最大价值=到原点的最大价值+原点的价值
         printf("%d\n",ans);
     }
     return ;
 }