ZOJ 3946.Highway Project(The 13th Zhejiang Provincial Collegiate Programming Contest.K) SPFA

ZOJ Problem Set - 3946

Highway Project

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 C_i dollars. It takes D_i minutes to travel between city X_i and Y_i 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 X_i, Y_i, D_i, C_i (0 ≤ X_i, Y_i < N, 0 < D_i, C_i < 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

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3946

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题意：求最短时间和最少花费。先满足最短时间，再满足最少花费。

思路：n,m很大，要用SPFA算法，不能用邻接表，否则会爆内存。

代码:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int MAXN = 1e5+100, mod = 1e9 + 7, inf = 0x3f3f3f3f;
typedef long long ll;
const ll INF = (1ll<<50);
struct node
{
    int to,d,c;
} edge[2*MAXN];
int head[MAXN],next[2*MAXN];
int sign[MAXN];
queue<int>Q;
ll dist[MAXN],cost[MAXN];
int n,m;
void add(int i,int u,int v,int d,int c)
{
    edge[i].to=v;
    edge[i].d=d;
    edge[i].c=c;
    next[i]=head[u];
    head[u]=i;
}
void SPFA(int v)
{
    int i,u;
    for(i=0; i<n; i++)
    {
        dist[i]=INF;cost[i]=INF;
        sign[i]=0;
    }
    dist[v]=0;
    cost[0]=0;
    Q.push(v);
    sign[v]=1;
    while(!Q.empty())
    {
        u=Q.front();
        Q.pop();
        sign[u]=0;
        i=head[u];
        while(i!=-1)
        {
            if(dist[edge[i].to]>dist[u]+edge[i].d)
            {
                dist[edge[i].to]=dist[u]+edge[i].d;
                cost[edge[i].to]=edge[i].c;
                if(!sign[edge[i].to])
                {
                    Q.push(edge[i].to);
                    sign[edge[i].to]=1;
                }
            }
            else if(dist[edge[i].to]==dist[u]+edge[i].d)
            {
                if(cost[edge[i].to]>edge[i].c)
                    cost[edge[i].to]=edge[i].c;
            }
            i=next[i];
        }
    }
}
int main()
{
    int i,j,T;
    int x,y,d,c;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        memset(edge,0,sizeof(edge));
        memset(next,0,sizeof(next));
        for(i=0; i<=n; i++) head[i]=-1;
        j=1;
        for(i=1; i<=m; i++)
        {
            scanf("%d%d%d%d",&x,&y,&d,&c);
            add(j,x,y,d,c);j++;
            add(j,y,x,d,c);j++;
        }
        SPFA(0);
        ll ans1=0,ans2=0;
        for(i=1; i<n; i++)
        {
            if(dist[i]!=INF) ans1+=dist[i];
            if(cost[i]!=INF) ans2+=cost[i];
        }
        cout<<ans1<<" "<<ans2<<endl;
    }
    return 0;
}