Sicily 1823 Hardest Job Ever!(单源最短路)

//单源最短路径//Dijkstra + 优先队列实现#include<iostream>#include<queue>#include<cstring>#define INF 1200000000#define MAXN 205#define MAXM 20005using namespace std;typedef pair<int,int> dis_v;//用一个pair类型存放当前结点的距离和结点编号int T,N,M;int head[MAXN],next[MAXM],V[MAXM],W[MAXM],dis[MAXN];int m;void addEdge(int u,int v,int w){V[m] = v;W[m] = w;next[m] = head[u];head[u] = m++;}void buildGraph(){int u,v,w;m = 0;memset(head,-1,sizeof(head));scanf("%d%d",&N,&M);while(M--){scanf("%d%d%d",&u,&v,&w);addEdge(u,v,w);addEdge(v,u,w);}}int Dijkstra(int st,int ed)//Dijkstra + 优先队列{for(int i = 1;i <= N;++i)dis[i] = i == st ? 0 : INF;priority_queue<dis_v> q;q.push(make_pair(dis[st],st));while(!q.empty()){dis_v x = q.top();q.pop();int u = x.second;if(x.first != dis[u])continue;//x.first为入队前dis[u]的距离，如果队列中的dis[u]和当前的dis[u]不同，证明u这个结点被处理过，而dijkstra算法每个节点只会被处理一次，所以跳过for(int e = head[u];e != -1;e = next[e])if(dis[u] + W[e] < dis[V[e]]){dis[V[e]] = dis[u] + W[e];q.push(make_pair(dis[V[e]],V[e]));}}if(dis[ed] == INF)return -1;//不可达else return dis[ed];}int main(){scanf("%d",&T);while(T--){buildGraph();printf("%d/n",Dijkstra(1,N));}return 0;}