UVa 10986 - Sending email (Dijkstra优化, SPFA)

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1927

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.

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 InputSample Output

32 1 0 10 1 1003 3 2 00 1 1000 2 2001 2 502 0 0 1

Case #1: 100Case #2: 150Case #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\n", d[end]);        else puts("unreachable");    }    return 0;}

2.SPFA,  0.160s

#include<cstdio>#include<cstring>#include<utility>#include<queue>using namespace std;queue<int>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 SPFA(int src){    memset(vis, 0, sizeof(vis));    for(int i=0; i<n; ++i) d[i] = INF;    d[src] = 0;    q.push(src);    while(!q.empty()){        int u = q.front();  q.pop();        vis[u] = false;        for(int e=head[u]; e!=-1; e=next[e])if(d[v[e]] > d[u]+w[e]){            d[v[e]] = d[u] + w[e];            if(!vis[v[e]]){                vis[v[e]] = true;                q.push(v[e]);            }        }    }}int main(){    int T,cas=1;    scanf("%d",&T);    while(T--){        read_graph();        SPFA(beg);        printf("Case #%d: ",cas++);        if(d[end]!=INF) printf("%d\n", d[end]);        else puts("unreachable");    }    return 0;}

——  生命的意义，在于赋予它意义。
          
     原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)