zoj2008 最短路

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題意：給你n個點的有向圖，從1點到其他所有點又從其他點回到1點的最短路。

 

思路：可以求一次從1點出發的最短路，再反向建圖，再求一次從1出發的最短路，把兩次的結果加起來就是題目所求。由於邊比較多，

  所以最好用Dijkstra+優先隊列，或者SPFA；

 

 

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>const int maxn = 1000009;const int inf = 1<<30;#include<vector>struct node{int v,w;node(int v,int w):v(v),w(w){}bool operator < (const node a)const{return w > a.w;}};std::vector<node> eg[maxn];std::priority_queue<node> q;int n,m;int vis[maxn];int dis[maxn];int a[maxn],b[maxn],c[maxn];void DIJ(){int i;memset(vis,0,sizeof(vis));for(i=0;i<=n;i++)dis[i] = inf;while(!q.empty())q.pop();dis[1] = 0;vis[1] = 0;int size = eg[1].size();for(i=0; i < size; i++){q.push(node(eg[1][i].v,eg[1][i].w));dis[eg[1][i].v] = eg[1][i].w;}while(!q.empty()){int u = q.top().v;int w = q.top().w;q.pop();if(!vis[u]){vis[u] = 1;size = eg[u].size();for(i=0; i<size; i++){int v = eg[u][i].v;int w = eg[u][i].w;if(!vis[v] && dis[u] + w < dis[v]){dis[v] = dis[u] + w;q.push(node(v,dis[v]));}}}}}void work(){int i,sum = 0;DIJ();for(i=2; i<=n; i++)   sum += dis[i]; //printf("%d ",dis[i]);}for(i=0; i<=n; i++) eg[i].clear();for(i=0;i<m;i++)    eg[b[i]].push_back(node(a[i],c[i]));  //  printf("\n");DIJ();for(i=2; i <= n;i++) sum += dis[i]; //printf("%d ",dis[i]);}printf("%d\n",sum);}void input(){int i;scanf("%d%d",&n,&m);for(i=0 ;i < m; i++)  scanf("%d%d%d",&a[i],&b[i],&c[i]);for(i=0;i<=n;i++)     eg[i].clear();for(i=0;i<m;i++)      eg[a[i]].push_back(node(b[i],c[i]));}int main(){int t;scanf("%d",&t);while(t--){input();work();}return 0;}/*32 21 2 132 1 334 61 2 102 1 601 3 203 4 102 4 54 1 502 21 2 132 1 33  */