poj3255

本文地址：http://www.javashuo.com/article/p-myjcygrr-ew.html

題目名稱：Roadblocks

連接：http://poj.org/problem?id=3255

題意：有 N 個點，R條邊，計算從 1 出發到 N的次短路長度，次短路是指長度比那些最短路長的路徑。

思路：emmm，最短路是每次保存最短的點，那咱們每次保存最短和次短就行了

代碼以下：

 

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<queue>
 4 using namespace std;  5 typedef pair<int,int> P;  6 const int INF = 0x3f3f3f3f;  7 const int maxn = 5005;  8 struct Edge{  9     int v, w; 10     int nxt; 11 }edge[100005 * 2]; 12 int head[maxn]; 13 int n, r; 14 int cnt, dist1[maxn], dist2[maxn]; 15 void add_edge(int u, int v, int w){ 16     edge[cnt].v = v; 17     edge[cnt].w = w; 18     edge[cnt].nxt = head[u]; 19     head[u] = cnt++; 20 } 21 void init(){ 22     cnt = 0; 23     fill(head, head + n + 1, -1); 24     fill(dist1, dist1 + n + 1, INF); 25     fill(dist2, dist2 + n + 1, INF); 26 } 27 void spfa(int s){ 28     priority_queue<P, vector<P>, greater<P> > Q; 29     dist1[s] = 0; 30     Q.push(P(0, 1)); 31     while(!Q.empty()){ 32         P p = Q.top(); Q.pop(); 33         int v = p.second, w = p.first; 34         if(dist2[v] < w) continue; 35         for(int i = head[v]; i != -1; i = edge[i].nxt){ 36             Edge &e = edge[i]; 37             int d = w + e.w; 38             if(dist1[e.v] > d){ 39  swap(dist1[e.v], d); 40  Q.push(P(dist1[e.v], e.v)); 41  } 42             if(dist2[e.v] > d && dist1[e.v] < d){ 43                 dist2[e.v] = d; 44  Q.push(P(dist2[e.v], e.v)); 45  } 46  } 47  } 48 } 49 int main(){ 50 
51     scanf("%d%d", &n, &r); 52  init(); 53     for(int i = 1; i <= r; i++){ 54         int a,b,d; 55         scanf("%d%d%d", &a, &b, &d); 56  add_edge(a, b, d); 57  add_edge(b, a, d); 58  } 59     spfa(1); 60     printf("%d\n", dist2[n]); 61     return 0; 62 }

View Code