【次短路】POJ3255 Roadblocks

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20419		Accepted: 7157

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)

Source

USACO 2006 November Gold

---

思路：次短路其实就是在最短路的基础上再添进来一条边使得起点到x的最短路+vis[x][y]+y至终点的最短路 最小

跑两遍最短路，再枚举一个边即可

---

#include <iostream>
#include <cstring>
//#pragma GCC optimize(2)
#include<time.h>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 5005
#define inf 1e18
#define eps 0.00001
typedef long long ll;
const ll mod = 1e9+7;
const double pi = acos(-1);

ll n,r,ans1[maxn],ans2[maxn];
vector < pair< ll, ll> > arr[maxn];
bool flag[maxn];

struct Why
{
    ll now,dis;
    Why(ll a, ll b)
    {
        now = a;
        dis = b;
    }
    bool operator < (const Why a) const
    {
        return dis > a.dis;
    }

};

void slove(ll x,ll *ans)
{
    priority_queue < Why> A;
    A.push( Why(x,0)  );

    while(!A.empty())
    {
        Why temp = A.top();
        A.pop();

        if( flag[temp.now] == 0 )
        {
            flag[temp.now] = 1;

            ans[ temp.now  ] = min(ans[temp.now],temp.dis );

            for(ll i = 0; i < arr[temp.now].size(); i++)
            {
                A.push( Why( arr[temp.now][i].first, temp.dis+arr[temp.now][i].second  )  );
            }

        }

    }

    return ;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);

    //freopen("D:\\test1.in","w",stdout);
    //srand((int)time(0));

    cin >> n >> r;

    memset(ans1,0x3f3f,sizeof(ans1));
    memset(ans2,0x3f3f,sizeof(ans2));

    while(r--)
    {
        ll a,b,c;
        cin >> a >> b >> c;
        arr[a].push_back( make_pair(b,c) );
        arr[b].push_back( make_pair(a,c) );
    }

    slove(1,ans1);
    memset(flag,0,sizeof(flag));
    slove(n,ans2);

    ll ans = inf;

    for(ll i = 1; i <= n; i++)
    {
        for(ll k = 0; k < arr[i].size(); k++)
        {
            if( ans1[i]+ans2[ arr[i][k].first ]+ arr[i][k].second > ans1[n])
                ans = min(ans,ans1[i]+ans2[ arr[i][k].first ]+ arr[i][k].second);
        }
    }
    
    cout << ans << endl;
    
    return 0;
}