ZOJ 3946 Highway Project(Dijkstra)
ZOJ 3946 Highway Project(Dijkstra)
Highway Project
Time Limit: 2 Seconds Memory Limit: 65536 KB
Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus, he proposed the highway project.
The Marjar Empire has N cities (including the capital), indexed from 0 to N - 1 (the capital is 0) and there are M highways can be built. Building the i-th highway costs Ci dollars. It takes Di minutes to travel between cityXi and Yi on the i-th highway.
Edward wants to find a construction plan with minimal total time needed to reach other cities from the capital, i.e. the sum of minimal time needed to travel from the capital to city i (1 ≤ i ≤ N). Among all feasible plans, Edward wants to select the plan with minimal cost. Please help him to finish this task.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first contains two integers N, M (1 ≤ N, M ≤ 105).
Then followed by M lines, each line contains four integers Xi, Yi, Di, Ci (0 ≤ Xi, Yi < N, 0 < Di, Ci < 105).
Output
For each test case, output two integers indicating the minimal total time and the minimal cost for the highway project when the total time is minimized.
Sample Input
2
4 5
0 3 1 1
0 1 1 1
0 2 10 10
2 1 1 1
2 3 1 2
4 5
0 3 1 1
0 1 1 1
0 2 10 10
2 1 2 1
2 3 1 2Sample Output
4 3
4 4
Dijkstra 用优先队列 ，用邻接表建立图，注意要long long int。比赛最后一分钟发现数组没有long long int. 还有在更新s数组的要加等号，#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <string>
#include <stdio.h>
#include <queue>
using namespace std;
const long long int INF=1000000000000000;
#define MAX 100000
struct Node
{
int value;
int next;
long long int time;
long long int cost;
}edge[MAX*2+5];
struct Node2
{
int time;
int cost;
int value;
Node2(){};
Node2(int time,int cost,int value)
{
this->time=time;
this->cost=cost;
this->value=value;
}
friend bool operator<(Node2 a,Node2 b)
{
if(a.time==b.time)
return a.cost>b.cost;
return a.time>b.time;
}
};
int cot;
int head[MAX+5];
int vis[MAX+5];
long long int s[MAX+5];
long long int ans1;
long long int ans2;
int n,m;
void add(int x,int y,long long int time,long long int cost)
{
edge[cot].value=y;
edge[cot].next=head[x];
edge[cot].time=time;
edge[cot].cost=cost;
head[x]=cot++;
}
void Dijkstra()
{
priority_queue<Node2> q;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
s[i]=INF;
s[0]=0;
q.push(Node2(0,0,0));
while(!q.empty())
{
Node2 term=q.top();
q.pop();
if(vis[term.value]) continue;
vis[term.value]=1;
ans2+=term.cost;
int a=head[term.value];
while(a!=-1)
{
if(s[edge[a].value]>=s[term.value]+edge[a].time)
{
s[edge[a].value]=s[term.value]+edge[a].time;
q.push(Node2(s[edge[a].value],edge[a].cost,edge[a].value));
}
a=edge[a].next;
}
}
}
int main()
{
int t;
scanf("%d",&t);
int x,y;
long long int xx,yy;
while(t--)
{
scanf("%d%d",&n,&m);
memset(head,-1,sizeof(head));
cot=0;
for(int i=1;i<=m;i++)
{
scanf("%d%d%lld%lld",&x,&y,&xx,&yy);
add(x,y,xx,yy);
add(y,x,xx,yy);
}
ans1=0;ans2=0;
Dijkstra();
for(int i=1;i<n;i++)
ans1+=s[i];
printf("%lld %lld\n",ans1,ans2);
}
return 0;
}腾讯云开发者
