L2-001. 紧急救援 Dijkstra

L2-001. 紧急救援

链接：

https://www.patest.cn/contests/gplt/L2-001

题解：

用一遍dijkstra算法。设立num[i]和w[i]表示从出发点到i结点拥有的路的条数，以及能够找到的救援队的数目~~~当判定dis[u] + e[u][v] < dis[v]的时候，不仅仅要更新dis[v]，还要更新num[v] = num[u], w[v] = weight[v] + w[u]; 如果dis[u] + e[u][v] == dis[v]，还要更新num[v] += num[u]，而且判断一下是否权重w[v]更小，如果更小了就更新w[v] = weight[v] + w[u]; 
再设立一个pre[i]表示最短路径的前一个结点，在dis[u] + e[u][v] <= dis[v]的时候更新pre[v] = u，最后递归打印路径即可

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-7;
// head

const int maxn = 510;
int cost[maxn][maxn];
int d[maxn];
int used[maxn];
int pre[maxn];
int V;
int N, M, S, D;
int weight[maxn], num[maxn], w[maxn];

void dijkstra(int s) {
    memset(d, 0x3f, sizeof(d));
    memset(used, 0, sizeof(used));
    memset(pre, -1, sizeof(pre));
    d[s] = 0;
    w[s] = weight[s];
    num[s] = 1;

    while (1) {
        int v = -1;
        rep(u, 0, N - 1) if (!used[u] && (v == -1 || d[u] < d[v])) v = u;
        if (v == -1) break; used[v] = 1;
        rep(u, 0, N - 1) if (used[u] == false && cost[u][v] != inf) {
            if (d[u] > d[v] + cost[v][u]) {
                d[u] = d[v] + cost[v][u];
                pre[u] = v;
                num[u] = num[v];
                w[u] = w[v] + weight[u];
            }
            else if (d[u] == d[v] + cost[v][u]) {
                num[u] += num[v];
                if (w[u] < w[v] + weight[u]) {
                    w[u] = w[v] + weight[u];
                    pre[u] = v;
                }
            }
        }
    }
}

VI get_path(int t) {
    VI path;
    for (; t != -1; t = pre[t]) path.pb(t);
    reverse(all(path));
    return path;
}

int main() {
    cin >> N >> M >> S >> D;
    V = N;
    rep(i, 0, N - 1) cin >> weight[i];
    memset(cost, 0x3f, sizeof(cost));
    while (M--) {
        int a, b, c;
        cin >> a >> b >> c;
        cost[a][b] = cost[b][a] = c;
    }
    dijkstra(S);
    cout << num[D] << " " << w[D] << endl;
    VI path = get_path(D);
    cout << path[0];
    rep(i, 1, path.size() - 1) cout << ' ' << path[i];
    cout << endl;
    return 0;
}

注意：

题目的序号是从0开始的，还有一定要把cost数组初始化为inf