ZOJ 3946 Highway Project（最短路）
题目链接：http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3946
题面：
Highway Project Time Limit: 2 Seconds Memory Limit: 65536 KBEdward, 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 4Author: Lu, Yi
Source: The 13th Zhejiang Provincial Collegiate Programming Contest
题目大意：
求从0点出发单独到其他点的最小时间代价，并求得加的边代价最小。
解题：
spfa加边，若当前点加入点使代价发生变化，且该点不在队列中，则将该点加入队列，并更新最小值。倘若该点时间代价与原有值相同，则比较到该点对应边的代价，若新代价小，则更新新的代价。若时间代价比原有大，则不更新。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define inf 0x3f3f3f3f
#define LL long long
using namespace std;
int head[100010],cnt,next[200010];
LL cost[100010],dist[100010];
bool vis[100010];
struct edge
{
int fm,to;
LL ti,cost;
}E[200010];
void addedge(int u,int v,LL ti,LL cos)
{
next[cnt]=head[u];
head[u]=cnt;
E[cnt].fm=u;
E[cnt].to=v;
E[cnt].ti=ti;
E[cnt].cost=cos;
cnt++;
}
queue <int> q;
int main()
{
int n,m,t,fm,to,d,c,tmp;
LL timecost,moneycost,val;
scanf("%d",&t);
while(t--)
{
timecost=moneycost=0;
while(!q.empty())
q.pop();
cnt=0;
memset(head,-1,sizeof(head));
memset(next,-1,sizeof(next));
memset(cost,inf,sizeof(inf));
memset(vis,0,sizeof(vis));
memset(dist,inf,sizeof(dist));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d%lld%lld",&fm,&to,&d,&c);
addedge(fm,to,d,c);
addedge(to,fm,d,c);
}
q.push(0);
vis[0]=1;
dist[0]=0;
while(!q.empty())
{
tmp=q.front();
q.pop();
vis[tmp]=0;
for(int i=head[tmp];~i;i=next[i])
{
val=dist[E[i].fm]+E[i].ti;
if(val<dist[E[i].to])
{
dist[E[i].to]=val;
if(!vis[E[i].to])
{
q.push(E[i].to);
vis[E[i].to]=1;
}
cost[E[i].to]=E[i].cost;
}
else if(val==dist[E[i].to])
{
if(cost[E[i].to]>E[i].cost)
cost[E[i].to]=E[i].cost;
}
else
continue;
}
}
for(int i=1;i<n;i++)
{
moneycost+=cost[i];
timecost+=dist[i];
}
printf("%lld %lld\n",timecost,moneycost);
}
return 0;
}