PAT 甲级 1003Emergency(Dijkstra最短路)

最新推荐文章于 2022-05-04 20:11:19 发布

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最新推荐文章于 2022-05-04 20:11:19 发布

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原文链接：http://www.cnblogs.com/dacc123/p/8228626.html

本文探讨了作为城市应急救援队领导时，如何利用特殊地图上的城市分布、连接道路及其长度，迅速响应紧急求助并最大化召集救援资源的策略。通过输入案例分析，展示了解决此类问题的算法实现，包括寻找从特定城市到另一特定城市的最短路径数量，以及在此过程中可能收集的最大救援资源量。

1003. Emergency (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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, C1 and C2 - 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 c1, c2 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 C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, 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

Sample Output

2 4

题目的意思要求求出最短路的个数，和点的权值最大的值

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <queue>

using namespace std;
const int maxn=1e9;
struct Nod
{
  int value;
  int next;
  int weight;
}edge[505*2];
int head[505];
int tot;
void add(int x,int y,int z)
{
  edge[tot].value=y;
  edge[tot].weight=z;
  edge[tot].next=head[x];
  head[x]=tot++;
}
int vis[505];
int d[505];
int num[505];
int res[505];
int w[505];
int n,c1,c2,m;
struct Node
{
  int pos;
  int dis;
  Node(){};
  Node(int pos,int dis)
  {
    this->pos=pos;
    this->dis=dis;
  }
  friend bool operator <(Node a,Node b)
  {
    return a.dis>b.dis;
  }
};
void Dijkstra(int st)
{
  priority_queue<Node> q;
  q.push(Node(st,0));
  memset(vis,0,sizeof(vis));
  memset(num,0,sizeof(num));
  memset(res,0,sizeof(res));
  for(int i=0;i<=500;i++)
    d[i]=1e9;
  d[st]=0;num[st]=1;res[st]=w[st];
  while(!q.empty())
  {
    Node term=q.top();
    q.pop();
    if(vis[term.pos])
      continue;
    vis[term.pos]=1;
    for(int i=head[term.pos];i!=-1;i=edge[i].next)
    {
      int y=edge[i].value;
      if(d[y]>term.dis+edge[i].weight)
      {
        num[y]+=num[term.pos];
        res[y]=res[term.pos]+w[y];
        d[y]=term.dis+edge[i].weight;
        q.push(Node(edge[i].value,d[y]));
      }
      else if(d[y]==term.dis+edge[i].weight)
      {
                num[y]+=num[term.pos];
        res[y]=max(res[y],res[term.pos]+w[y]);
      }
    }
  }
}

int main()
{
  scanf("%d%d%d%d",&n,&m,&c1,&c2);
  for(int i=0;i<n;i++)
    scanf("%d",&w[i]);
  int x,y,z;
  memset(head,-1,sizeof(head));
  tot=0;
  for(int i=1;i<=m;i++)
  {
    scanf("%d%d%d",&x,&y,&z);
    add(x,y,z);
    add(y,x,z);
  }
  Dijkstra(c1);
  printf("%d %d\n",num[c2],res[c2]);
  return 0;
}

转载于:https://www.cnblogs.com/dacc123/p/8228626.html