文章标题

ZOJ-3946 Highway Project

思路：简单的单源最短路径（dijkstra模板），dij[i]数组用于记录源点到城市i的最小time，cost[i]记录从某一前置城市到达城市i的最小花费（满足time的前提）；更新dij[i]时必须更新cost[i]，遇到与当前dij[i]相同时，更新cost[i]更新为较小值。

题目链接

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 998244353
#define eps 1e-6
double pi=acos(-1);
const ll INF=1e12;
#define MAXN 100009
struct node{
int city;
ll time,cost;
node(int c,ll t,ll co){
city=c;time=t;cost=co;
}
bool operator < (const node &a) const {
return time > a.time;//最小值优先
}
};
ll dij[MAXN],cost[MAXN];//time,cost
int n;
bool flag[MAXN];
vector<node> ma[MAXN];//邻接表
priority_queue<node> que;//优先队列
void dijkstra(int sourse){
int i,u,v;
ll w,c;
for(i=0;i<n;i++){
dij[i]=(i==0?0:INF);//初始化
cost[i]=(i==0?0:INF);
}
memset(flag,false,sizeof(flag));
while(!que.empty()){
que.pop();
}
que.push(node(0,0,0));
while(!que.empty()){
node pa=que.top();
que.pop();
u=pa.city;
if(flag[u]){
continue;
}
flag[u]=true;
int len=ma[u].size();
for(i=0;i<len;i++){
v=ma[u][i].city;
w=ma[u][i].time;
c=ma[u][i].cost;
if(dij[v]>dij[u]+w){
dij[v]=dij[u]+w;
cost[v]=c;
if(!flag[v]){
que.push(node(v,dij[v],c));
}

}
else if(dij[v]==dij[u]+w){
if(cost[v]>c){
cost[v]=c;
}
}
}
}
ll ret=0,Cost=0;
for(i=0;i<n;i++){
ret+=dij[i];
Cost+=cost[i];
}
printf("%lld %lld\n",ret,Cost);
return;
}

int main()
{

int T,m,i;
int u,v;
ll w,c;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)ma[i].clear();
for(i=0;i<m;i++){
scanf("%d%d%lld%lld",&u,&v,&w,&c);
ma[u].push_back(node(v,w,c));
ma[v].push_back(node(u,w,c));
}
dijkstra(0);
}
return 0;
}