UVa 10986：Sending email （Dijkstra优化, SPFA）

UVa 10986：Sending email （Dijkstra优化, SPFA）2014-10-16 csdn博客 shuangde800链接：

题目：

Problem E
Sending email
Time Limit: 3 seconds

"A new internet watchdog is creating a stir in
Springfield. Mr. X, if that is his real name, has
come up with a sensational scoop."
Kent Brockman

There are n SMTP servers connected by network cables. Each of the m cables connects two computers and has a certain latency measured in milliseconds required to send an email message. What is the shortest time required to send a message from server S to server T along a sequence of cables？ Assume that there is no delay incurred at any of the servers.

URL：http://www.bianceng.cn/Programming/sjjg/201410/45900.htm

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing n（2<=n<20000）, m （0<=m<50000）, S （0<=S<n） and T （0<=T<n）. S！=T. The next m lines will each contain 3 integers: 2 different servers （in the range [0, n-1]） that are connected by a bidirectional cable and the latency, w, along this cable （0<=w<=10000）.

Output
For each test case, output the line "Case #x:" followed by the number of milliseconds required to send a message from S toT. Print "unreachable" if there is no route from S to T.
Sample Input3
2 1 0 1
0 1 100
3 3 2 0
0 1 100
0 2 200
1 2 50
2 0 0 1

Sample Output
Case #1: 100
Case #2: 150
Case #3: unreachable

Problemsetter: Igor Naverniouk

题目大意:

给一个图，求从s点到t点的最小距离。

分析与总结：

赤裸裸的最短路，但n太大显然是不能用邻接矩阵的，需要用邻接表+优先队列优化。

代码:

1. Dijkstra, 0.148s

#include<cstdio>#include<cstring>#include<utility>#include<queue>using namespace std;typedef pair<int,int>pii;priority_queue<pii,vector<pii>,greater<pii> >q;const int N = 100005;const int INF = 1000000000;int n, m, beg, end, k;int head[N], next[N], u[N], v[N], w[N], d[N];bool vis[N]; inline void read_graph（）{scanf（"%d%d%d%d",&n,&m,&beg,&end）;memset（head, -1, sizeof（head））;for（int e=1; e<=m; ++e）{scanf（"%d%d%d",&u[e],&v[e],&w[e]）;u[e+m]=v[e], v[e+m]=u[e], w[e+m]=w[e];next[e] = head[u[e]];head[u[e]] = e;next[e+m] = head[u[e+m]];head[u[e+m]] = e+m;}}inline void Dijkstra（int src）{memset（vis, 0, sizeof（vis））;for（int i=0; i<n; ++i） d[i] = INF;d[src] = 0;q.push（make_pair（d[src], src））;while（！q.empty（））{pii u = q.top（）;q.pop（）;int x = u.second;if（vis[x]） continue;vis[x] = true;for（int e=head[x]; e！=-1; e=next[e]）if（d[v[e]] > d[x]+w[e]）{d[v[e]] = d[x]+w[e];q.push（make_pair（d[v[e]], v[e]））;}}}int main（）{int T,cas=1;scanf（"%d",&T）;while（T--）{read_graph（）;Dijkstra（beg）;printf（"Case #%d: ",cas++）;if（d[end]！=INF） printf（"%d
", d[end]）;else puts（"unreachable"）;}return 0;}