13th省赛K

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 city Xi 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 2

Sample Output

4 3
4 4

题目大意：

问从0节点到所有节点的最短步数，如果步数相同，那么输出步数最短的并且花费最少的路径

思路

dijskstra就好了。（表示inf设置小了。。。卡了我三个小时。。。）

#include

using namespace std;

typedef long long ll;
const ll inf = 1e18;
const int maxn = 100000 + 100;
struct edge{
int u, v;
ll d, c;
edge(int u, int v, ll d, ll c): u(u), v(v), d(d), c(c){}
};
vector  e;
vector  G[maxn];
struct node{//c是价格，d是时间
int u;
ll d, c ;
node (int u, ll d, ll c): u(u), d(d), c(c){}
bool operator < (const node &a) const{
if (a.d != d) return d > a.d;
return c > a.c;
}
};
ll d[maxn];
ll cost[maxn];
int n, m;
bool vis[maxn];

void init(){
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++){
int u, v;
ll d, c;
cin >> u >> v >> d >> c;
e.push_back(edge(u, v, d, c));
e.push_back(edge(v, u, d, c));
int l = e.size();
G[u].push_back(l - 2);
G[v].push_back(l - 1);
}
}

void solve(){
for (int i = 0; i <= n; i++){
d[i] = inf;
cost[i] = inf;
vis[i] = false;
}
priority_queue  q;
d[0] = 0;
cost[0] = 0;
q.push(node(0, d[0], cost[0]));
while (!q.empty()){
node x = q.top();
q.pop();
int u = x.u;//现在所处的顶点
if (vis[u] == true) continue;
vis[u] = true;
for (int i = 0; i < G[u].size(); i++){
edge y = e[G[u][i]];
if (d[y.v] > d[u] + y.d){
cost[y.v] = y.c;
d[y.v] = y.d + d[u];
q.push(node(y.v, d[y.v], cost[y.v]));
}
else if (d[y.v] == d[u] + y.d && cost[y.v] > y.c){
cost[y.v] = y.c;
q.push(node(y.v, d[y.v], cost[y.v]));
}
}
}
ll resc = 0, resd = 0;
for (int i = 1; i < n; i++){
resc += cost[i];
resd += d[i];
}
cout << resd << " " << resc << endl;
for (int i = 0; i < n; i++){
G[i].erase(G[i].begin(), G[i].end());
}
e.erase(e.begin(), e.end());
}

int main(){
int t;
scanf("%d", &t);
while (t--){
init();
solve();
}
return 0;
}