单源次短路径：poj：3255-Roadblocks

Roadblocks

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 17521 Accepted: 6167

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

Hint

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

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解题心得：

1. 提议就是给你一个无向图，叫你求出第1点到第n点的次短路径。
2. 最短路径很好求，方法也很多，其实次短路径也很好求，就拿dij来说，每次维护的都是一个最小值，那么次短路径就可以在最小值之上维护一个和最小值差值最小的值。
3. 到某个顶点v的次短路要么是到其他某个顶点u的最短路加上u到v的边，要么是到u的次短路加上u到v的边。因此所需要求的就是到所有顶点的最短路和次短路。

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#include <stdio.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
typedef pair<int,int> P;
const int maxn = 5010;

vector <P> ve[maxn];
int n,m,dis[maxn],dis2[maxn];
priority_queue <P ,vector<P>,greater<P> > qu;

void init() {
    memset(dis,0x3f,sizeof(dis));
    memset(dis2,0x3f,sizeof(dis2));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int u,v,va;
        scanf("%d%d%d",&u,&v,&va);
        ve[u].push_back(make_pair(va,v));
        ve[v].push_back(make_pair(va,u));
    }
    dis[1] = 0;
    qu.push(make_pair(0,1));
}

void dij() {
    while(!qu.empty()) {
        P now = qu.top();
        qu.pop();
        int u = now.second;
        int d = now.first;
        if(dis2[u] < d)
            continue;
        for(int i=0;i<ve[u].size();i++){
            int v = ve[u][i].second;
            int d2 = d + ve[u][i].first;
            if(d2 < dis[v]) {
                swap(dis[v],d2);
                qu.push(make_pair(dis[v],v));
            }
            if(d2 > dis[v] && d2 < dis2[v]) {
                dis2[v] = d2;
                qu.push(make_pair(d2,v));
            }
        }
    }
    return ;
}

int main() {
    init();
    dij();
    printf("%d
",dis2[n]);
    return 0;
}