POJ--3013[Big Christmas Tree] （SPFA）

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题意:

重点理解：price of an edge will be (sum of weights of all descendant nodes) × (unit price of the edge)；看似求类似最小生成树的题，其实仔细想想可以转换成最短路的问题;

即：给你一幅无向图，求root点到每个点的最短距离(dis[i])*每个点的权值(W[i])的和;price of an edge will be (sum of weights of all descendant nodes) × (unit price of the edge)

源代码：

/*SPFA*//*思路:ans=root到每个结点的最短距离(dis[i])*每个点的权值(W[i])的和注意点：(1):无向图(2):用__int64*//*AC代码：672ms*/#include <iostream>#include <cstdio>#include <memory.h>#include <queue>#include <algorithm>#define MAXN 50005#define min(a,b) (a<b?a:b)#define max(a,b) (a>b?a:b)#define INF 1e18using namespace std;struct edge{int u,v,next;__int64 w;}E[2*MAXN];int head[MAXN],ecnt;__int64 W[MAXN];__int64 dis[MAXN];bool vis[MAXN];int N,M;void Insert(int u,int v,__int64 w){E[ecnt].u=u;E[ecnt].v=v;E[ecnt].w=w;E[ecnt].next=head[u];head[u]=ecnt++;}void Init(){int i,u,v;__int64 w;memset(head,-1,sizeof(head));ecnt=0;scanf("%d%d",&N,&M);for(i=1;i<=N;i++)scanf("%I64d",&W[i]);for(i=1;i<=M;i++){scanf("%d%d%I64d",&u,&v,&w);Insert(u,v,w);Insert(v,u,w);}}queue<int>Q;void SPFA(){int i,u,v;__int64 w;while(!Q.empty()) Q.pop();memset(vis,false,sizeof(vis));for(i=1;i<=N;i++)dis[i]=INF;dis[1]=0;vis[1]=true;Q.push(1);while(!Q.empty()){u=Q.front();Q.pop();vis[u]=false;for(i=head[u];i!=-1;i=E[i].next){v=E[i].v;w=E[i].w;if(dis[v]>dis[u]+w){dis[v]=dis[u]+w;if(!vis[v]){vis[v]=true;Q.push(v);}}}}}void Solve(){int i;SPFA();__int64 ans=0;bool ok=true;for(i=1;i<=N;i++){if(dis[i]==INF){ok=false;break;}ans+=dis[i]*W[i];}if(!ok)printf("No Answer\n");elseprintf("%I64d\n",ans);}int main(){int T;scanf("%d",&T);while(T--){Init();Solve();}return 0;}