pat-a1003. Emergency (25)

这个题本身是一个比较简单的单源最短路径的问题。也不怕被笑这个算法一直写得有问题。这次仔细学习了大佬的代码。要不断重复才能熟悉这个算法。

#include<cstdio>
#include<map>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
int we[510];
int visit[510];
int dis[510];
int ans[510];
int c2;
struct node{
int to,p,cs;
node(int a=0,int b=0,int c=0):to(a),p(b),cs(c){
}
friend bool operator < (const node& a,const node& b){
if(a.p!=b.p) return a.p>b.p;
return a.cs<b.cs;
}
};
map<int,vector<node> >g;
int cnt[510];
void dij(int s){
dis[s]=0;
cnt[s]=1;
priority_queue<node> q;
ans[s]=we[s];
q.push({s,0,ans[s]});
while(!q.empty()){
node t=q.top();
q.pop();
int f=t.to;
if(visit[f]) continue;
visit[f]=1;
int len=g[f].size();
for(int i=0;i<len;++i){
node e=g[f][i];
if(dis[e.to]>dis[f]+e.p){
cnt[e.to]=cnt[f];
}
else if(dis[e.to]==dis[f]+e.p){
cnt[e.to]+=cnt[f];
}
if(dis[e.to]>dis[f]+e.p||(dis[e.to]==dis[f]+e.p&&ans[e.to]<ans[f]+we[e.to])){
dis[e.to]=dis[f]+e.p;
ans[e.to]=ans[f]+we[e.to];
q.push({e.to,dis[e.to],ans[e.to]});

}
}
}
}
int main(){
int n,m,c1;
int a,b,c;
scanf("%d%d%d%d",&n,&m,&c1,&c2);
for(int i=0;i<n;++i) scanf("%d",we+i);
for(int i=0;i<n;++i) dis[i]=100000000;
while(m--){
scanf("%d%d%d",&a,&b,&c);
g[a].push_back(node(b,c,we));
g[b].push_back(node(a,c,we[a]));
}
dij(c1);
printf("%d %d\n",cnt[c2],ans[c2]);
return 0;
}

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.

[b]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

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