poj 3255 Roadblocks（次最短路径）

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10836		Accepted: 3852

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)

题目大意：找到从1->N的次最短路径

次最短路的求解：

到某个顶点的次最短路

1、到其他某个顶点u的最短路＋u->v的边

2、到某个定点的次最短路＋u->v的边

不断更新最短／次最短路的数组

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <algorithm>
#include <map>
#include <math.h>
#include <stack>
#include <queue>
#define Max 6000
#define inf 100000000
using namespace std;
int N,R;
struct edge{
    int to;
    int cost;
};
vector<edge> G[Max];
int dis[Max];
int dis2[Max];
typedef pair<int, int> P;
void solve(){
    priority_queue<P,vector<P>,greater<P> > q;
    fill(dis, dis+N, inf);
    fill(dis2, dis2+N, inf);
    dis[0] = 0;
    q.push(P(0,0));
    while (!q.empty()) {
        P p = q.top();q.pop();
        int v = p.second, d = p.first;
        if (dis2[v]<d) {
            continue;
        }
        for (int i=0; i<G[v].size(); i++) {
            edge &e = G[v][i];
            int d2 = e.cost + d;
            if (dis[e.to]>d2) {
                swap(dis[e.to], d2);
                q.push(P(dis[e.to],e.to));
            }
            if (dis2[e.to]>d2&&dis[e.to]<d2) {
                dis2[e.to] = d2;
                q.push(P(dis2[e.to],e.to));
            }
        }
    }
    printf("%d\n",dis2[N-1]);
}
int main(){
    while (scanf("%d%d",&N,&R)!=EOF) {
        for (int i=0; i<=N; i++) {
            G[i].clear();
        }
        for (int i=0; i<R; i++) {
            int s,e,c;
            scanf("%d%d%d",&s,&e,&c);
            s--,e--;
            edge p;
            p.to = e,p.cost = c;
            G[s].push_back(p);
            p.to = s;
            G[e].push_back(p);
        }
        solve();
    }
    return 0;
}