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最新推荐文章于 2023-07-12 17:13:31 发布

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最新推荐文章于 2023-07-12 17:13:31 发布

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原文链接：http://www.cnblogs.com/Stephen-F/p/9908897.html

本文详细介绍了如何使用两遍最短路径算法解决特定问题，通过正向和反向计算，有效地找到从1号点到所有其他点的最短路径，并计算总距离。文章深入解析了算法的核心思想，提供了完整的C++实现代码，包括读取输入、建立图结构、执行Dijkstra算法等关键步骤。

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核心思想:两遍最短路.

1号点去各地的时间直接套最短路模板,各地到1号点时间用逆向思维,视为求1号点沿反边到各地的时间即可.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define re register
using namespace std;
const int maxn = 1000005;

inline int read(){
    char ch = getchar();
    int f = 1 , x = 0;
    while(ch > '9' || ch < '0'){if(ch == '-')f = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + ch - '0';ch = getchar();}
    return x * f;
}

int n,m,u,v;
long long w;
int head1[maxn],tot1,head2[maxn],tot2;
long long ans,dis1[maxn],dis2[maxn];

struct Edge{
    int from,to,next;
    long long val;
}edge1[maxn<<1] , edge2[maxn<<1];

struct Node{
    int u,d;
    bool operator < (const Node &f) const {
        return d > f.d;
    }
};

inline void add1(int u,int v,long long w){
    edge1[++tot1].from = u;
    edge1[tot1].to = v;
    edge1[tot1].val = w;
    edge1[tot1].next = head1[u];
    head1[u] = tot1;
}

inline void add2(int u,int v,long long w){
    edge2[++tot2].from = u;
    edge2[tot2].to = v;
    edge2[tot2].val = w;
    edge2[tot2].next = head2[u];
    head2[u] = tot2;
}

inline void dijk1(int s){
    for(re int i=1;i<=n;i++)  dis1[i] = 1e9;
    priority_queue<Node> q;
    dis1[s] = 0;
    q.push((Node) {s , dis1[s]});
    while(!q.empty()){
        Node cur = q.top();
        q.pop();
        int d = cur.d , u = cur.u;
        if(d != dis1[u]) continue;
        for(re int i=head1[u];i;i=edge1[i].next){
            int v = edge1[i].to;
            if(dis1[v] > dis1[u] + edge1[i].val){
                dis1[v] = dis1[u] + edge1[i].val;
                q.push((Node) {v , dis1[v]});
            }
        }
    }
}

inline void dijk2(int s){
    for(re int i=1;i<=n;i++)  dis2[i] = 1e9;
    priority_queue<Node> q;
    dis2[s] = 0;
    q.push((Node) {s , dis2[s]});
    while(!q.empty()){
        Node cur = q.top();
        q.pop();
        int d = cur.d , u = cur.u;
        if(d != dis2[u]) continue;
        for(re int i=head2[u];i;i=edge2[i].next){
            int v = edge2[i].to;
            if(dis2[v] > dis2[u] + edge2[i].val){
                dis2[v] = dis2[u] + edge2[i].val;
                q.push((Node) {v , dis2[v]});
            }
        }
    }
}

int main(){
    n = read(); m = read();
    for(int i=1;i<=m;i++){
        u = read(); v = read(); scanf("%lld",&w);
        add1(u , v , w);
        add2(v , u , w);
    }
    dijk1(1);
    for(int i=2;i<=n;i++)
        ans += dis1[i];
    dijk2(1);   
    for(int i=n;i>=2;i--){  
        ans += dis2[i];
    }
    printf("%lld\n",ans);
    return 0;
}

转载于:https://www.cnblogs.com/Stephen-F/p/9908897.html