ZOJ 3946-Highway Project【最短路的应用】（2016浙江省大学生程序设计竞赛）

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 toN - 1 (the capital is 0) and there areM highways can be built. Building thei-th highway costs
Ci dollars. It takes Di minutes to travel between cityXi andYi 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 cityi (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 2

Sample Output

4 3
4 4
解题思路：
给出一张图，边无向有两个，两个权分别为路的消耗的时间和花费的经前，点编号0-n-1，现在要你求从0建立到其他的路径，使各点花费总时间最少优先的情况下花费的金钱也最少。就是要考虑都是最短的情况下，谁的花费最少。
[code]#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf=0x3f3f3f3f;
const ll maxn=100005;
ll edgenum,vis[maxn],head[maxn],dist[maxn],cost[maxn];
struct node
{
ll from,to,len,val;
ll next;
}edge[maxn*5];
void init()
{
edgenum=0;
memset(head,-1,sizeof(head));
memset(dist,inf,sizeof(dist));
memset(vis,0,sizeof(vis));
memset(cost,inf,sizeof(cost));
}
void add(ll u,ll v,ll d,ll c)
{
node tp={u,v,d,c,head[u]};
edge[edgenum]=tp;
head[u]=edgenum++;
}
void dijkstra(ll n,ll st)
{
queue<ll> q;
q.push(st);
vis[st]=1;
dist[st]=0;
while(!q.empty())
{
ll v=q.front();q.pop();
vis[v]=0;
for(ll i=head[v];i!=-1;i=edge[i].next)
{
ll u=edge[i].to;
if(dist[edge[i].to]==dist[v]+edge[i].len&&edge[i].val<cost[edge[i].to])
{
cost[edge[i].to]=edge[i].val;
}
if(dist[edge[i].to]>dist[v]+edge[i].len)
{
dist[edge[i].to]=dist[v]+edge[i].len;
cost[edge[i].to]=edge[i].val;
if(!vis[edge[i].to])
{
vis[edge[i].to]=1;
q.push(edge[i].to);
}
}
}
}
}
int main()
{
//freopen("shuju.txt","r",stdin);
ll t;
scanf("%lld",&t);
while(t--)
{
ll n,m;
scanf("%lld%lld",&n,&m);
init();
for(ll i=0;i<m;++i)
{
ll u,v,d,c;
scanf("%lld%lld%lld%lld",&u,&v,&d,&c);
add(u,v,d,c);add(v,u,d,c);
}
ll minlenth=0,mincost=0;
dijkstra(n,0);
for(ll i=1;i<n;++i)
{
minlenth+=dist[i];
mincost+=cost[i];
}
printf("%lld %lld\n",minlenth,mincost);
}
return 0;
}

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