201712-4 CCF 行车路线堆优化dijkstra_ccf2017行车路线-优快云博客

本文链接：https://blog.youkuaiyun.com/jay__bryant/article/details/82709850

本文介绍了一种基于Dijkstra算法的优化方法，通过维护以i结尾的连续小路的长度和，实现对最短路径计算的高效处理。适用于带权图的最短路径求解，特别关注了边权重对路径总成本的影响。

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维护一个sum[]：最短路径中以i结尾的连续小路的长度和

#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;
typedef long long ll;
const ll INF = 1e18;
const int N = 510;
const int M = 2e5+10;

struct node
{
    int p;
    ll w;
    node(){}
    node(int _p, ll _w):p(_p),w(_w){}
    bool operator <(const node &a)const
    {
        return w>a.w;
    }
};
bool vis[N];
ll dis[N];
ll sum[N];//最短路径中以i结尾的连续小路的长度和
int to[M],type[M],nxt[M];
ll val[M];
int head[N],tot;
int n, m;

void addedge(int u, int v, int t, int w)
{
    ++tot;
    to[tot] = v;
    type[tot] = t;
    val[tot] = w;
    nxt[tot] = head[u];
    head[u] = tot;
}

void dijkstra(int s)
{
    priority_queue <node> q;
    memset(vis, 0, sizeof(vis));
    memset(sum, 0, sizeof(sum));
    for(int i = 1; i <= n; ++i) dis[i] = INF;
    dis[s] = 0;
    q.push(node(s, 0));
    while(!q.empty())
    {
        node u = q.top();
        q.pop();
        int p = u.p;
        if(vis[p]) continue;
        vis[p] = 1;
        for(int i = head[p]; ~i; i = nxt[i])
        {
            int v = to[i];
            if(vis[v]) continue;
            ll cost = 0;
            if(type[i])
            {
                ll tmp = sum[p]+val[i];
                cost = dis[p]-sum[p]*sum[p]+tmp*tmp;
            }
            else cost = dis[p]+val[i];
            if(cost<dis[v])
            {
                dis[v] = cost;
                if(type[i])
                    sum[v] = sum[p]+val[i];
                else sum[v] = 0;
                q.push(node(v, dis[v]));
            }
        }
    }
}

int main()
{
    scanf("%d %d", &n, &m);
    memset(head, -1, sizeof(head));
    tot = 0;
    while(m--)
    {
        int t, u, v, w;
        scanf("%d %d %d %d", &t, &u, &v, &w);
        addedge(u, v, t, w);
        addedge(v, u, t, w);
    }
    dijkstra(1);
    printf("%lld\n", dis[n]);
    return 0;
}