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, 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 Specification:

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

题目大意:求最短路的数量和最短路经过的最大点权

思路:改造dijkstra算法,

• sumw[i] 表示到达节点i的最短路的最大点权和
• num[i] 表示到达节点i的最短路的数量
• 初始化时,都赋值为0
• 对于start节点,sumw就是其点权,num就是1

核心松弛操作代码如下:如果小于的话,直接把sum加上那个点权,然后num就是原来的值,

如果相等的话,num加上另一路最短路的num值,sum如果也比他大就更新

if(!vis[v]&&dis[v]>dis[u]+cost)
{
    dis[v]=dis[u]+cost;
    sumw[v]=sumw[u]+weight[v];
    num[v]=num[u];///有说道
    Q.push(qnode(v,dis[v]));
}
else if(!vis[v]&&dis[v]==dis[u]+cost)
{
    num[v]+=num[u];
    if(sumw[u]+weight[v]>sumw[v])
    {
        sumw[v]=sumw[u]+weight[v];
    }
    Q.push(qnode(v,dis[v]));
}

完整AC代码:

#include<bits/stdc++.h>
#define de(x) cout<<#x<<" "<<(x)<<endl
#define each(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
const int maxn=500+5;
const int inf=0x3f3f3f3f;
int dis[maxn];
int weight[maxn];
int sumw[maxn];
int num[maxn];
bool vis[maxn];
struct Edge
{
    int v,c;
    Edge(int v,int c):v(v),c(c){}
};
vector<Edge>G[maxn];
struct qnode
{
    int v,c;
    qnode(int v=0,int c=0):v(v),c(c){}
    bool operator<(const qnode&r)const
    {
        return c>r.c;
    }
};
void Dijkstra(int n,int start)
{
    memset(vis,0,sizeof(vis));
    for(int i=0;i<=n;i++)
    {
        dis[i]=inf;
        sumw[i]=0;
        num[i]=0;
    }
    priority_queue<qnode>Q;
    while(!Q.empty())Q.pop();
    dis[start]=0;
    sumw[start]=weight[start];
    num[start]=1;
    Q.push(qnode(start,0));///居然忘了push进去了
    qnode temp;
    while(!Q.empty())
    {
        temp=Q.top();
        Q.pop();
        int u=temp.v;
        if(vis[u])continue;
        vis[u]=true;
        for(int i=0;i<G[u].size();i++)
        {
            int v=G[u][i].v;
            int cost=G[u][i].c;
            if(!vis[v]&&dis[v]>dis[u]+cost)
            {
                dis[v]=dis[u]+cost;
                sumw[v]=sumw[u]+weight[v];
                num[v]=num[u];///有说道
                Q.push(qnode(v,dis[v]));
            }
            else if(!vis[v]&&dis[v]==dis[u]+cost)
            {
                num[v]+=num[u];
                if(sumw[u]+weight[v]>sumw[v])
                {
                    sumw[v]=sumw[u]+weight[v];
                }
                Q.push(qnode(v,dis[v]));
            }
        }
    }

}
/*
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
*/
int main()
{
    int n,m;
    int from,to;
    cin>>n>>m>>from>>to;
    each(i,0,n-1)cin>>weight[i];
    each(i,1,m)
    {
        int a,b,c;
        cin>>a>>b>>c;
        G[a].push_back(Edge(b,c));
        G[b].push_back(Edge(a,c));
    }
    Dijkstra(n,from);
    printf("%d %d\n",num[to],sumw[to]);
    return 0;
}