POJ--3255 Roadblocks(最短路)

记录
0:25 2023-1-27

http://poj.org/problem?id=3255

reference:《挑战程序设计竞赛(第2版)》2.4.4 p108

Description

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

Input

Line 1: Two space-separated integers: N and R
Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

4 4
1 2 100
2 4 200
2 3 250
3 4 100

Sample Output

450

最短路，求最短路径和次短路径，其实相当于维护了一个次短距离。(说实话，看得我有点浮躁，以前看算法还是比较平淡的，现在可能焦躁了起来) 还有一件事，好久不做算法题感觉大脑都要锈了。急?寄!
这道题还要注意是双向的，所以读入边在邻接表表示中要注意输入俩次。

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<vector>
#include<utility>
#include<queue>
using namespace std;
#define MAX_N 100000

const int INF = 0x7fffffff;
int N, R;

struct edge{int to, cost;};
typedef pair<int, int> P;  //first 最短距离 second顶点编号

vector<edge> G[MAX_N + 1];
int dist[MAX_N + 1];
int dist2[MAX_N + 1];

void solve() {
    priority_queue<P, vector<P>, greater<P> > que;
    fill(dist, dist + N, INF);
    fill(dist2, dist2 + N, INF);
    dist[0] = 0;
    que.push(P(0, 0));

    while (!que.empty()) {
        P p = que.top(); que.pop();
        int v = p.second, d = p.first;
        if(dist2[v] < d) continue;
        for(int i = 0; i < G[v].size(); i++) {
            edge e = G[v][i];
            int d2 = d + e.cost;
            if(dist[e.to] > d2) {
                swap(dist[e.to], d2);
                que.push(P(dist[e.to], e.to));
            }
            if(dist2[e.to] > d2 && dist[e.to] < d2) {
                dist2[e.to] = d2;
                que.push(P(dist2[e.to], e.to));
            }
        }
    }
    printf("%d\n", dist2[N - 1]);
}

int main() {
    scanf("%d %d", &N, &R);
    int from, to, cost;
    edge e;
    for (int i = 0; i < R; i++) {
        scanf("%d %d %d", &from, &to, &cost);
        e.to = to - 1;
        e.cost = cost;
        G[from - 1].push_back(e);
        e.to = from - 1;
        G[to - 1].push_back(e);
    }
    solve();
}