CCF认证 2016-09 交通规划

题目链接：http://118.190.20.162/view.page?gpid=T44

读懂题意后是要求在单源最短路的基础上找总长最小的路径

在Dijkstra算法的基础上再加一些判断花费最小的语句就可以了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<queue>using namespace std;const int N=1e4+10;const int inf=1e8-1;struct edge{int to;int cost;edge(){}edge(int a,int b){to=a;cost=b;}};struct node{int id;int cost;node(int a,int b){id=a;cost=b;}bool operator < (const node &u) const{return cost>u.cost;}};vector<edge> g[N];int dist[N],f[N]; //f[]数组存储到每个节点的单条路径花费int n,m;void init(){for(int i=0;i<=n;i++)dist[i]=f[i]=inf;dist[1]=0;}void dj(){priority_queue<node> q;q.push(node(1,0));while(!q.empty()){node x=q.top();q.pop();int u=x.id;int t=x.cost;int l=g[u].size();for(int i=0;i<l;i++){if(dist[g[u][i].to]>dist[u]+g[u][i].cost||dist[g[u][i].to]==dist[u]+g[u][i].cost&&f[g[u][i].to]>g[u][i].cost) //更短路或者是同短路加花费更小{dist[g[u][i].to]=dist[u]+g[u][i].cost;q.push(node(g[u][i].to,dist[g[u][i].to]));f[g[u][i].to]=g[u][i].cost; //更新花费}}}}int main(){int a,b,c;scanf("%d%d",&n,&m);for(int i=0;i<m;i++){scanf("%d%d%d",&a,&b,&c);g[a].push_back(edge(b,c));g[b].push_back(edge(a,c));}init();dj();int ans=0;for(int i=2;i<=n;i++)ans+=f[i];printf("%d\n",ans);return 0;}