ZOJ3946：Highway Project（最短路变形）
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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[sub]i[/sub] dollars. It takes D[sub]i[/sub] minutes to travel between city X[sub]i[/sub] and Y[sub]i[/sub] 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 ≤ 10[sup]5[/sup]).
Then followed by M lines, each line contains four integers X[sub]i[/sub], Y[sub]i[/sub], D[sub]i[/sub], C[sub]i[/sub] (0 ≤ X[sub]i[/sub], Y[sub]i[/sub] < N, 0 < D[sub]i[/sub], C[sub]i[/sub] < 10[sup]5[/sup]).<h4< dd=””>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.<h4< dd=””>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
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<h4< dd=””>Sample Output
- 4 34 4
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题意：给一个图，每条边有个距离和花费，要求创建一个子图，满足0点到其余点的距离总和最小，且边的总花费最小。
思路：条件一，每个点贪心的都选最短路就是最优解，条件二，类似最小生成树的kruskal算法，在条件一的基础上更改下松弛条件即可。
拓展：CF545E，求距离总和最小，且总的边的长度最小，其实那题也是类似最短路基础上跑最小生成树。
- # include <iostream># include <cstdio># include <cstring># include <queue>using namespace std;const int maxn = 2e5+30;long long dis[maxn], cost[maxn], x, y;int vis[maxn], Next[maxn], cnt;struct node{int u, v, w, c, next;}edge[maxn];queue<int>q;int scan(){ int res=0,ch,flag=0; if((ch=getchar())==”-“) flag=1; else if(ch>=”0″&&ch<=”9”) res=ch-“0″; while((ch=getchar())>=”0″&&ch<=”9”) res=res*10+ch-“0”; return flag?-res:res;}void out(long long x){ if(x/10) out(x/10); putchar(“0″+x%10);}void add(int u, int v, int w, int c){ edge[cnt] = {u, v,w,c,Next[u]};; Next[u] = cnt++; edge[cnt] = {v, u,w,c,Next[v]}; Next[v] = cnt++;}void spfa(){ memset(dis, 0x3f, sizeof(dis)); memset(cost, 0x3f, sizeof(cost)); memset(vis, 0, sizeof(vis)); while(!q.empty()) q.pop(); q.push(1); dis[1] = cost[1] = 0; while(!q.empty()) { int u = q.front(); q.pop(); vis[u] = 0; for(int i=Next[u]; ~i; i=edge[i].next) { int v=edge[i].v, w=edge[i].w, c=edge[i].c; if(dis[v] > dis[u]+w) { dis[v] = dis[u]+w; cost[v] = c; if(!vis[v]) { vis[v] = 1; q.push(v); } } else if(dis[v] == dis[u]+w) { if(cost[v] > c) { cost[v] = c; if(!vis[v]) { vis[v] = 1; q.push(v); } } } } }}int main(){ int t, n, m, u, v, w, c; t=scan(); while(t–) { x = y = cnt = 0; memset(Next, -1, sizeof(Next)); n=scan();m=scan(); for(int i=1; i<=m; ++i) { u=scan();v=scan();w=scan();c=scan(); add(u+1,v+1,w,c); } spfa(); for(int i=2; i<=n; ++i) { x += dis[i]; y += cost[i]; } out(x); putchar(” “); out(y); puts(“”); } return 0;}
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