ZOJ Problem Set - 3946 Highway Project

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 toN - 1 (the capital is 0) and there are M highways can be built. Building thei-th highway costs C_i dollars. It takes D_i minutes to travel between cityX_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

24 50 3 1 10 1 1 10 2 10 102 1 1 12 3 1 24 50 3 1 10 1 1 10 2 10 102 1 2 12 3 1 2

Sample Output

4 34 4一道裸的spfa。做题的时候没有将cost和i赋值为足够大，一直wa，烦死了
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>using namespace std;typedef long long LL;const int N = 50005 * 5;const LL INF = 100000000000;struct Node{LL v, c, d;int next;}node[N];int head[N];int pos = 0;LL cost[N], T[N];bool inq[N];void init(){pos = 0;fill(cost, cost + N, INF);fill(T, T + N, INF);memset(inq, 0, sizeof(inq));memset(head, -1, sizeof(head));}void addEdge(int u, int v, int t, int c){//前向星node[pos].v = v;node[pos].c = c;node[pos].d = t;node[pos].next = head[u];head[u] = pos++;}void SPFA(){queue<int> q;q.push(0);inq[0] = 1;cost[0] = 0;T[0] = 0;while (!q.empty()){int v = q.front();//循环到节点vq.pop();inq[v] = 0;for (int i = head[v]; i + 1; i = node[i].next){int pos = node[i].v;//城市v能到达的城市if (T[v] + node[i].d <= T[pos]){if (T[v] + node[i].d < T[pos]){T[pos] = T[v] + node[i].d;cost[pos] = node[i].c;if (!inq[pos]){//城市pos是否在队列中inq[pos] = 1;q.push(pos);}}else if (cost[pos]>node[i].c){cost[pos] = node[i].c;if (!inq[pos]){inq[pos] = 1;q.push(pos);}}}}}}int main(){int t;#ifdef glxfreopen("in.txt", "r", stdin);#endifscanf("%d", &t);while (t--){int n, m;scanf("%d%d", &n, &m);int u, v, c, d;init();for (int i = 0; i < m; i++){scanf("%d%d%d%d", &u, &v, &d, &c);addEdge(u, v, d, c);addEdge(v, u, d, c);}SPFA();LL dd = 0, cc = 0;for (int i = 0; i < n; i++){dd += T[i];cc += cost[i];}printf("%lld %lld\n", dd, cc);}return 0;}