poj 3255

POJ - 3255点我点我:-)


Roadblocks

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

SubmitStatus

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 intersectionN.

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 andR
Lines 2.. R+1: Each line contains three space-separated integers: A,B, andD that describe a road that connects intersectionsA andB and has lengthD (1 ≤ D ≤ 5000)

Output

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

Sample Input

4 41 2 1002 4 2002 3 2503 4 100

Sample Output

450

求次短路。。。dijstra算法好。。。短。。。

这个做法就是多加了个d2[]，求次短路。。

k短。。。不是很清楚，本星期搞定吧

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>#include<queue>using namespace std;#define MAXN (5000+5)#define INF 0x3f3f3f3ftypedef pair<int, int> P;struct node{    int v, dis;};int n;vector<node> G[MAXN];int d[MAXN], d2[MAXN];void dijstra(int s){     priority_queue<P, vector<P>, greater<P> > q;     for(int i = 1; i <= n; i++) d[i] = d2[i] = INF;          d[s] = 0;     q.push(P(d[s], s));          while(!q.empty()){         P now = q.top(); q.pop();         int u = now.second, dis = now.first;         if(d2[u] < dis) continue;                  for(int i = 0; i < G[u].size(); i++){             int v = G[u][i].v, c2 = dis+G[u][i].dis;                          if(d[v] > c2){                 swap(d[v], c2);                 q.push(P(d[v], v));             }             if(d2[v] > c2 && d[v] < c2){                 d2[v] = c2;                 q.push(P(d2[v], v));             }         }     }}int main(){    freopen("test.in", "r", stdin);    freopen("test.out", "w", stdout);        int m;    scanf("%d%d", &n, &m);        for(int i = 1; i <= m; i++){        int u, v, dis;        scanf("%d%d%d", &u, &v, &dis);                G[u].push_back((node){v, dis});        G[v].push_back((node){u, dis});    }        dijstra(1);        printf("%d\n", d2[n]);        return 0;}

ＣＯＭＥ　ＯＮ！