aoj-2249 Road Construction 单源最短路dijkstra+堆优化（模板）

来源：互联网发布：java开发一年工作经验编辑：程序博客网时间：2024/05/16 14:04

先用dijkstra求出capital到其他的各个点的最短距离，然后最后遍历一遍各个边，把之前用到的最短距离的边的最小消耗加在一起即可。（详见代码）

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cstdlib>#include<cmath>#include<set>#include<bitset>#include<map>#include<stack>#include<queue>#include<vector>#include<utility>#define INF 0x3f3f3f3f#define inf 100000000000000000using namespace std;typedef long long ll;typedef pair<int,int>P;int n,m,d[10005],u,v,dd,cc;struct edge{    int f,t,d,c;    edge(int f,int t,int d,int c):f(f),t(t),d(d),c(c){};};vector<edge>G[10005];void dijkstra(int s){    priority_queue<P,vector<P>,greater<P> >que;    que.push(P(0,s));    memset(d,INF,sizeof(d));d[s]=0;    while(!que.empty())    {        P p=que.top();que.pop();        if(d[p.second]<p.first)continue;        for(int i=0;i<G[p.second].size();i++)        {            if(d[G[p.second][i].t]>d[p.second]+G[p.second][i].d)            {                d[G[p.second][i].t]=d[p.second]+G[p.second][i].d;                que.push(P(d[G[p.second][i].t],G[p.second][i].t));            }        }    }}int main(){    while(scanf("%d%d",&n,&m)!=EOF&&(n!=0||m!=0))    {        for(int i=1;i<=n;i++)G[i].clear();        for(int i=0;i<m;i++)        {            scanf("%d%d%d%d",&u,&v,&dd,&cc);            G[u].push_back(edge(u,v,dd,cc));            G[v].push_back(edge(v,u,dd,cc));        }         dijkstra(1);         int ans=0;         for(int i=2;i<=n;i++)         {             int temp=INF;             for(int j=0;j<G[i].size();j++)             {                 if(d[G[i][j].t]+G[i][j].d==d[i])temp=min(temp,G[i][j].c);             }             ans+=temp;         }         printf("%d\n",ans);    }    return 0;}

0 0