[BZOJ 2100] Apple Delivery

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=2100

[算法]

        Answer = min{ dist(PB,PA1) + dist(PA1,PA2) , dist(PB,PA2) + dist(PA1,PA2) } （其中，dist表示最短路)

        对PB和PA1分别求两次最短路即可，注意要使用dijkstra算法（堆优化） 

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXP 100010
#define MAXC 200010

struct edge
{
        int to,w,nxt;
} e[MAXC << 1];

int i,C,P,PA1,PA2,PB,u,v,w,tot;
int head[MAXP],dist1[MAXP],dist2[MAXP];

inline void addedge(int u,int v,int w)
{
        tot++;
        e[tot] = (edge){v,w,head[u]};
        head[u] = tot;
}
inline void dijkstra1()
{
        int i,cur,v,w;
        static bool visited[MAXP];
        priority_queue< pair<int,int> > q;
        memset(visited,false,sizeof(visited));
        memset(dist1,0x3f,sizeof(dist1));
        dist1[PB] = 0;
        q.push(make_pair(0,PB));
        while (!q.empty())
        {
                cur = q.top().second;
                q.pop();
                if (visited[cur]) continue;
                visited[cur] = true;
                for (i = head[cur]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;
                        if (dist1[cur] + w < dist1[v])
                        {
                                dist1[v] = dist1[cur] + w;
                                q.push(make_pair(-dist1[v],v));
                        }
                }
        }
}
inline void dijkstra2()
{
        int i,cur,v,w;
        static bool visited[MAXP];
        priority_queue< pair<int,int> > q;
        memset(visited,false,sizeof(visited));
        memset(dist2,0x3f,sizeof(dist1));
        dist2[PA1] = 0;
        q.push(make_pair(0,PA1));
        while (!q.empty())
        {
                cur = q.top().second;
                q.pop();
                if (visited[cur]) continue;
                visited[cur] = true;
                for (i = head[cur]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;
                        if (dist2[cur] + w < dist2[v])
                        {
                                dist2[v] = dist2[cur] + w;
                                q.push(make_pair(-dist2[v],v));
                        }
                }
        }
}
int main() 
{
        
        scanf("%d%d%d%d%d",&C,&P,&PB,&PA1,&PA2);
        for (i = 1; i <= C; i++)
        {
                scanf("%d%d%d",&u,&v,&w);
                addedge(u,v,w);
                addedge(v,u,w);        
        }
        dijkstra1();
        dijkstra2();
        printf("%d
",min(dist1[PA1] + dist2[PA2],dist1[PA2] + dist2[PA2]));
        
        return 0;
    
}