poj 3592 Instantaneous Transference tarjan缩点 最长路 ++tarjan模版 && spfa最长路模版

/*题意：给定一个矩阵，西南角为起点，每个单元都有一定价值的金矿（#表示岩石，不可达，*表示时空门，可以到达指定单元）      现在要求得最多可以获得多大利益题解：强联通分量，最长路；如果没有时空门，就是纯粹的有向无环图的最长路了，现在出现时空门了，只要求强联通分量进行       缩点，对缩点后的图建立有向无环图，然后求最长路就ok了*/#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<climits>#include<queue>#include<vector>using namespace std;const int MAXV=49*49;int n,m;int value[MAXV];char graph[41][41];struct spfa//  spfa求最长路{static const int INF =-(1<<30);struct Edge {int v,next;int w;} ep[MAXV*MAXV];int first[MAXV];int d[MAXV];int e;int V;inline void init(int n) {memset(first,-1,sizeof(first));e=0;V=n;}void addEdge(int u, int v, int dd) {ep[e].v=v;ep[e].w=dd;ep[e].next=first[u];first[u]=e++;}void solve(int source) {static bool ok[MAXV];memset(ok,false,sizeof(ok));queue<int> q;q.push(source);for (int i = 0; i < V; i++) d[i] = INF;d[source] = 0;while(!q.empty()){int x=q.front();q.pop();ok[x]=false;for(int k=first[x];k!=-1;k=ep[k].next){int v=ep[k].v;if(d[v]<d[x]+ep[k].w){d[v]=d[x]+ep[k].w;if(ok[v]==false){ok[v]=true;q.push(v);}}}}}} g;struct TarJan{struct Edge{int v,next;}edge[4*MAXV];int first[MAXV],ins[MAXV],dfn[MAXV],low[MAXV],Stack[MAXV],scc[MAXV];int top,index,sccnum,e,V;void init (int n){e=0;V=n;memset(first,-1,sizeof(first[0])*n);memset(ins,0,sizeof(ins[0])*n);memset(dfn,0,sizeof(dfn[0])*n);  memset(low,0,sizeof(dfn[0])*n);}void addEdge(int u,int v){ edge[e].v=v;edge[e].next=first[u];first[u]=e++;}void solve(){index=1;top=-1;sccnum=-1;for(int i=0;i<V;i++)if(dfn[i]==0) tarjan(i);}void tarjan(int u)  {  int v;   low[u] = dfn[u] = index++;   Stack[++top]=u; ins[u] = true;  for (int k=first[u]; k!=-1; k=edge[k].next)  {  v = edge[k].v;  if (dfn[v] == 0) {  tarjan(v);  low[u]=min(low[u],low[v]); }  else if (ins[v]) {  low[u]=min(low[u],dfn[v]);  }  }  if (dfn[u] == low[u])  {  sccnum++;  do{v = Stack[top--];ins[v] = false; scc[v] = sccnum;}while(u != v); }  }  }my;void init(){int x,y;memset(value,0,sizeof(value));my.init(n*m);for(int i=0;i<n;i++){for(int j=0;j<m;j++)if(graph[i][j]!='#'){if(i!=0)my.addEdge((i-1)*m+j,i*m+j);if(j!=0)my.addEdge(i*m+j-1,i*m+j);if(graph[i][j]=='*'){scanf("%d%d",&x,&y);if(graph[x][y]!='#')my.addEdge(i*m+j,x*m+y);}}}my.solve();}void solve(){g.init(my.sccnum+2);for(int i=0;i<n;i++)for(int j=0;j<m;j++){if(graph[i][j]!='*'&&graph[i][j]!='#'){value[my.scc[i*m+j]]+=graph[i][j]-'0';}}for(int i=0;i<n*m;i++){for(int k=my.first[i];k!=-1;k=my.edge[k].next){int v=my.edge[k].v;if(my.scc[i]!=my.scc[v]){g.addEdge(my.scc[i],my.scc[v],value[my.scc[v]]);}}}int ans=0;g.solve(my.scc[0]);for(int i=0;i<=my.sccnum;i++){if(g.d[i]>ans)ans=g.d[i];}printf("%d\n",ans+value[my.scc[0]]);}int main(){int ca;scanf("%d",&ca);while(ca--){scanf("%d%d",&n,&m);for(int i=0;i<n;i++){scanf("%s",graph[i]);}init();solve();}return 0;}