Emergency - C++

Emergency

1003
Emergency (25分)
As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.
Input Specification:
Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (≤500) - the number of cities (and the cities are numbered from 0 to N−1), M - the number of roads, C​1​​ and C​2​​ - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c​1​​, c​2​​ and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C​1​​ to C​2​​.
Output Specification:
For each test case, print in one line two numbers: the number of different shortest paths between C​1​​ and C​2​​, and the maximum amount of rescue teams you can possibly gather. All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.
Sample Input:
5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output:
2 4~

这是PTA甲级的一道题。难度不大。但是有些细节困扰了我很久。所以特此写一篇纪念一下。
话不多说，这道题就是求两点之间满足路径最短的路的条数并且顶点值和最多。我则想着随便用个什么求最短路径，再在路径最短的基础上，深搜即可。于是就用了spfa求最短路径。
C++代码如下：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct edge{
  int v,next;
}e[300000];
int n,m,sta,End,cnt; 
int node[510]; 
int s[300000];
int head[510];
int dis[510];
int book[510]; 
int Inf = 999999999;
queue<int> q;
int mins,maxw,count;
void add(int u,int v,int w)
{
  cnt ++;
  e[cnt].v = v;
  e[cnt].next = head[u];
  head[u] = cnt;
  s[cnt] = w;
} 
void spfa()
{
   //切记队列不为空时循环 
   while(q.size())
   {
     int k = q.front();
     book[k] = 0;
     q.pop();
     for(int i = head[k]; i; i = e[i].next)
     {
    int nextv = e[i].v;
    //只有 
      if(dis[nextv] > dis[k] + s[i])
 {
   dis[nextv] = dis[k] + s[i];
     if(book[nextv] == 0)
     {
    q.push(nextv);
    book[nextv] = 1; 
     }    
  }
     }  
   }
}
void dfs(int u, int w, int S)
{
    if(u == End)
    {
      if(S == mins)
      {
        count ++;
        maxw = max(maxw,w); 
  } 
      return;
    }
    for(int i = head[u]; i; i = e[i].next)
    {
       if(book[e[i].v] == 0)
       {
         int nextv = e[i].v;
         book[nextv] = 1;
         dfs(nextv,w + node[nextv],S + s[i]);
     book[nextv] = 0; 
   }
    }
}
int main()
{
   scanf("%d%d%d%d",&n,&m,&sta,&End);
   for(int i = 0; i < n; i ++)
     scanf("%d",&node[i]);
   for(int i = 1; i <= m; i ++)
   {
    int u,v,s;
    scanf("%d%d%d",&u,&v,&s);
    add(u,v,s);
    add(v,u,s);
   }
   for(int i = 0; i < n; i ++)
   dis[i] = Inf;
   dis[sta] = 0;
   //spfa求最短路径
   book[sta] = 1;
   q.push(sta);
   spfa(); 
   //求出最短路径再深搜
   memset(book,0,sizeof(book));
   book[sta] = 1;
   mins = dis[End];
   dfs(sta,node[sta],0);
   //求得是最短路径的条数不是求最短路径 
   cout << count << " " << maxw;
   return 0;
}

主要是以前没有注意的一点细节就是，spfa算法中一定要在松弛条件中判断是否入队列，并且每次出完队列后book数组要重新设置为0。