最小路径树 Codeforces545E Paths and Trees

传送门：点击打开链接

题意：告诉你图，和起点，求一棵树，使得起点走树上的路径到达其他的节点的路径是原来图中的最短的，且要求起点到所有点的距离之和最短。

思路：其实就是dijkstra的思路，只不过要稍微注意几个东西，首先是要保存上一条边的id，然后就是要注意当dist相同时，两条路径的长度相等时，应该要考虑最后一条边的长度，应该要选长度小的，因为两种方法实际上都是最短的，但是如果选边短的，就说路径上另一段是最长的，那么就有更多的公共部分了，这样会使得所有距离之和最短。

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;

const int MX = 3e5 + 5;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;

struct Edge {
    int v, nxt, cost;
} E[MX << 1];
int Head[MX], rear;
void edge_init() {
    rear = 0;
    memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v, int cost) {
    E[rear].v = v;
    E[rear].cost = cost;
    E[rear].nxt = Head[u];
    Head[u] = rear++;
}

struct Data {
    LL dist;
    int u, pre;
    Data(int _u, int _pre, LL _dist) {
        u = _u; pre = _pre; dist = _dist;
    }
    bool operator<(const Data &P) const {
        if(dist == P.dist){
            return E[pre].cost > E[P.pre].cost;
        }
        return dist > P.dist;
    }
};

bool vis[MX];
LL d[MX], ans;
vector<int>s;
void solve(int bx) {
    ans = 0; s.clear();
    memset(d, INF, sizeof(d));
    memset(vis, 0, sizeof(vis));

    d[bx] = 0;
    priority_queue<Data>Q;
    Q.push(Data(bx, -1, 0));
    while(!Q.empty()) {
        Data f = Q.top();
        Q.pop();

        if(vis[f.u]) continue;
        vis[f.u] = 1;
        if(f.pre != -1) {
            s.push_back(f.pre / 2 + 1);
            ans += E[f.pre].cost;
        }

        for(int i = Head[f.u]; ~i; i = E[i].nxt) {
            int v = E[i].v;
            if(f.dist + E[i].cost <= d[v]) {
                LL dist = f.dist + E[i].cost;
                Q.push(Data(v, i, dist));
                d[v] = dist;
            }
        }
    }

    printf("%I64d\n", ans);
    int siz = s.size();
    for(int i = 0; i < siz; i++) {
        printf("%d%c", s[i], i == siz - 1 ? '\n' : ' ');
    }
}

int main() {
    int n, m, bx; //FIN;
    while(~scanf("%d%d", &n, &m)) {
        edge_init();
        for(int i = 1; i <= m; i++) {
            int u, v, cost;
            scanf("%d%d%d", &u, &v, &cost);
            edge_add(u, v, cost);
            edge_add(v, u, cost);
        }
        scanf("%d", &bx);
        solve(bx);
    }
    return 0;
}