1030 Travel Plan (30分)（一个字短！）

1030 Travel Plan (30分)

分析

采用迪杰斯特拉算法，选取最短路径的同时记录前驱节点，若到同一点的最短路径不止一条，取花费最少的一条

代码

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
int n, m, s, d, a, b, l, co, distTo[510], pre[510], marked[510], cost[510];
vector<pair<int, pair<int, int> > >g[510];
struct node {
    int v, len;
    node() {}
    node(int vv, int ll) :v(vv), len(ll) {}
    bool operator<(node e)const {
        return v > e.v;
    }
};
int main() {
    scanf("%d %d %d %d", &n, &m, &s, &d);
    while (--m >= 0) {
        scanf("%d %d %d %d", &a, &b, &l, &co);
        g[a].push_back(pair<int, pair<int, int> >(b, pair<int, int>(l, co)));
        g[b].push_back(pair<int, pair<int, int> >(a, pair<int, int>(l, co)));
    }
    fill(distTo, distTo + 510, 999999999);
    fill(pre, pre + 510, -1);
    fill(cost, cost + 510, 999999999);
    priority_queue<node> q;
    q.push(node(s, 0));
    distTo[s] = 0;
    cost[s] = 0;
    while (!q.empty()) {
        int f = q.top().v;
        if (f == d)break;
        q.pop();
        marked[f] = 1;
        for (int i = 0; i < g[f].size(); ++i) {
            if (marked[g[f][i].first])continue;
            a = g[f][i].first, l = g[f][i].second.first, co = g[f][i].second.second;
            if (distTo[a] > distTo[f] + l) {
                distTo[a] = distTo[f] + l;
                cost[a] = cost[f] + co;
                pre[a] = f;
                q.push(node(a, distTo[a]));
            }
            else if (distTo[a] == distTo[f] + l) {
                if (cost[a] > cost[f] + co) {
                    cost[a] = cost[f] + co;
                    pre[a] = f;
                }
            }
        }
    }
    vector<int> o;
    a = d;
    while (d != -1) {
        o.insert(o.begin(), d);
        d = pre[d];
    }
    for (auto it : o)printf("%d ", it);
    printf("%d %d", distTo[a], cost[a]);
    return 0;
}