Aizu 2249 Road Construction【最短路变形】

Road Construction

Time Limit : 8 sec, Memory Limit : 65536 KB

Problem H: Road Construction

King Mercer is the king of ACM kingdom. There are one capital and some cities in his kingdom. Amazingly, there are no roads in the kingdom now. Recently, he planned to construct roads between the capital and the cities, but it turned out that the construction cost of his plan is much higher than expected.

In order to reduce the cost, he has decided to create a new construction plan by removing some roads from the original plan. However, he believes that a new plan should satisfy the following conditions:

• For every pair of cities, there is a route (a set of roads) connecting them.
• The minimum distance between the capital and each city does not change from his original plan.

Many plans may meet the conditions above, but King Mercer wants to know the plan with minimum cost. Your task is to write a program which reads his original plan and calculates the cost of a new plan with the minimum cost.

Input

The input consists of several datasets. Each dataset is formatted as follows.

N M
u1 v1 d1 c1 
.
.
.
uM vM dM cM 

The first line of each dataset begins with two integers, N and M (1 ≤ N ≤ 10000, 0 ≤ M ≤ 20000). N and Mindicate the number of cities and the number of roads in the original plan, respectively.

The following M lines describe the road information in the original plan. The i-th line contains four integers, ui,vi, di and ci (1 ≤ ui, vi ≤ N , ui ≠ vi , 1 ≤ di ≤ 1000, 1 ≤ ci ≤ 1000). ui , vi, di and ci indicate that there is a road which connects ui-th city and vi-th city, whose length is di and whose cost needed for construction is ci.

Each road is bidirectional. No two roads connect the same pair of cities. The 1-st city is the capital in the kingdom.

The end of the input is indicated by a line containing two zeros separated by a space. You should not process the line as a dataset.

Output

For each dataset, print the minimum cost of a plan which satisfies the conditions in a line.

Sample Input

3 31 2 1 22 3 2 13 1 3 25 51 2 2 22 3 1 11 4 1 14 5 1 15 3 1 15 101 2 32 101 3 43 431 4 12 521 5 84 232 3 58 422 4 86 992 5 57 833 4 11 323 5 75 214 5 23 435 101 2 1 531 3 1 651 4 1 241 5 1 762 3 1 192 4 1 462 5 1 253 4 1 133 5 1 654 5 1 340 0

Output for the Sample Input

35137218

题意：

给出若干个建筑之间的一些路，每条路都有对应的长度和需要的花费，问在保证源点1 到其他个点的距离最短的情况下，最少的花费是多少

题解：

首先明确需要使用邻接表存图，解题的第一个想法就是最短路上套一个最小生成树，但是相对会比较复杂，后来想，在更新最短路的时候，同时进行最小花费的更新，应该是行得通的，毕竟花费想当于是二级动态更新的属性，效果应该是一样的，然后上手做了.....

因个人对最短路其实并不是很了解，最终实现的时候程序存在漏洞，错了好几次才AC....

网上的题解都是用spfa 做的，说是这道题考的就是spfa ，个人不是太理解，spfs 和dijkstra  在处理不存在负权值的边时的更新操作难道不是一样的，都是进行松弛操作？

然后硬是用dijkstra 做了....

/*http://blog.csdn.net/liuke19950717*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int inf=0x3f3f3f3f;const int maxn=10005;int edgenum,head[maxn],dist[maxn];struct node{int from,to,len,val;int next;}edge[maxn*5];void init(int n){edgenum=0;memset(head,-1,sizeof(head));memset(dist,inf,sizeof(dist));}int add(int u,int v,int d,int c){node tp={u,v,d,c,head[u]};edge[edgenum]=tp;head[u]=edgenum++;}int dijkstra(int n,int st){int vis[maxn]={0},num[maxn];memset(num,-1,sizeof(num));//num 保存最后更新某个点的最短路使用了哪条边dist[st]=0;while(1){int v=-1;for(int u=1;u<=n;++u){if(!vis[u]&&(v==-1||dist[u]<dist[v])){v=u;}}if(v==-1){break;}vis[v]=1;for(int i=head[v];i!=-1;i=edge[i].next){int u=edge[i].to;if(dist[u]>dist[v]+edge[i].len){dist[u]=dist[v]+edge[i].len;num[u]=i;}else if(dist[u]==dist[v]+edge[i].len){if(num[u]!=-1&&edge[num[u]].val>edge[i].val)//考虑能否减少花费{num[u]=i;}}}}int ans=0;for(int i=1;i<=n;++i){if(num[i]!=-1){ans+=edge[num[i]].val;}}return ans;}int main(){int n,m;//freopen("shuju.txt","r",stdin);while(scanf("%d%d",&n,&m),n|m){init(n);for(int i=0;i<m;++i){int u,v,d,c;scanf("%d%d%d%d",&u,&v,&d,&c);add(u,v,d,c);add(v,u,d,c);}printf("%d\n",dijkstra(n,1));}return 0;}