題干：

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?Tindicating 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

題目大意：

給你一個無向圖，每條邊有一個距離c和花費d，叫你選一些邊，使得點0到其他所有點的距離之和最小，其次，使總花費最小。

解題報告：

? ?因為總題最優解一定滿足局部局部解，所以距離和最小就可以用求最短路來求解，最后統計個答案就可以。對于第二個權值的維護，可以在Dijkstra的同時貪心第二個權值，也可以先構造一個最短路子圖，求其中的最短路樹，也就是求最小樹形圖。（注意這里不是Kruskal可以解決的，，，因為其實這個新圖加的是有向邊、、、）

比如這個樣例：

不難看出，最短路子圖應該是這樣的：

而沒有由上到下的那條邊。

?

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; struct Edge {int fr,to;ll w,c;int ne; } e[MAX]; struct Graph {struct Point {int pos;ll c;Point() {}Point(int pos,ll c):pos(pos),c(c) {}bool operator<(const Point b) const {return c > b.c;}};Edge e[MAX];int tot = 0;//總共tot條邊，編號0~(tot-1)int head[MAX];bool vis[MAX];ll d[MAX],cost[MAX];void add(int u,int v,ll w,ll c) {e[tot].fr = u;e[tot].to = v;e[tot].ne = head[u];e[tot].w = w;e[tot].c = c;head[u] = tot++;}void Dijkstra(int st) {priority_queue<Point> pq;memset(d,0x3f,sizeof d);memset(cost,0x3f,sizeof cost);memset(vis,0,sizeof vis);d[st] = 0;cost[st] = 0;pq.push(Point(st,0));while(pq.size()) {Point cur = pq.top();pq.pop();if(vis[cur.pos]) continue;vis[cur.pos] = 1;for(int i = head[cur.pos]; ~i; i = e[i].ne) {if(d[e[i].to] > d[cur.pos] + e[i].w) {d[e[i].to] = d[cur.pos] + e[i].w;cost[e[i].to] = e[i].c;pq.push(Point(e[i].to,d[e[i].to]));}else if(d[e[i].to] == d[cur.pos]+e[i].w) {if(cost[e[i].to] > e[i].c) {cost[e[i].to] = e[i].c;pq.push(Point(e[i].to,d[e[i].to]));}}}}}void id_add(int u,int v,ll w,ll c) {e[tot].fr = u;e[tot].to = v;e[tot].ne = head[u];e[tot].w = w;e[tot].c = c;head[u] = tot++;}void init(int n) {tot = 0;for(int i = 0; i<=n; i++) head[i] = -1;} } G1; int main() {int n,m;int u,v;ll w,c;int t;cin>>t;while(t--) {scanf("%d%d",&n,&m);G1.init(n);for(int i = 1; i<=m; i++) {scanf("%d%d%lld%lld",&u,&v,&w,&c);u++,v++;G1.add(u,v,w,c);G1.add(v,u,w,c); }G1.Dijkstra(1);ll ans1 = 0,ans2 = 0;for(int i = 1; i<=n; i++) ans1 += G1.d[i],ans2 += G1.cost[i];printf("%lld %lld\n",ans1,ans2);}return 0 ; }

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總結

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