poj3255 Roadblocks

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13594		Accepted: 4783

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

Sample Output

Hint

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

Source

USACO 2006 November Gold

[Submit] [Go Back] [Status] [Discuss]

把求最短路的算法稍微改一下，每次记录节点的最短路和次短路。

在求最短路的时候会一次次地更新到那个结点的距离，在更新到真正的最短的距离时候，他的上一个就是次短距离，每次把那个值记录下来就行了。

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <map>
#define for(i,x,n) for(int i=x;i<n;i++)
#define ll long long int
#define INF 0x3f3f3f3f
#define MOD 1000000007

using namespace std;

int MAX_V,MAX_E;
struct edge{int to,cost;};
typedef pair<int,int> P;//最短距离，编号
vector<edge> G[5005];
int d[5005];//最短距离
int d2[5005];//次短距离

void dijkstra(int s){
    priority_queue<P,vector<P>,greater<P> > que;
    fill(d,d+MAX_V,INF);
    fill(d2,d2+MAX_V,INF);
    d[s]=0;
    que.push(P(d[s],s));
    while(!que.empty()){
        P p=que.top(); que.pop();
        int v=p.second;
        for(i,0,G[v].size()){
            int to=G[v][i].to;
            int dd=p.first+G[v][i].cost;
            if(dd<d[to]){
                d[to]^=dd;
                dd^=d[to];
                d[to]^=dd;
                que.push(P(d[to],to));
            }
            if(dd<d2[to] && dd>d[to]){
                d2[to]=dd;
                que.push(P(d2[to],to));
            }
        }
    }
}

int main()
{
    int x,y,z;
    scanf("%d %d",&MAX_V,&MAX_E);
    for(i,0,MAX_E){
        scanf("%d %d %d",&x,&y,&z);
        x-=1;
        y-=1;
        edge ee;
        ee.to=y; ee.cost=z;
        G[x].push_back(ee);
        ee.to=x; ee.cost=z;
        G[y].push_back(ee);
    }
    dijkstra(0);
    printf("%d",d2[MAX_V-1]);
    return 0;
}