UVa 10986 - Sending email (Dijkstra优化, SPFA)

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 Input Sample Output
3
2 1 0 1
0 1 100
3 3 2 0
0 1 100
0 2 200
1 2 50
2 0 0 1
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：
易做图裸的最短路，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
[cpp]
#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
[cpp]
#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("unr

补充：软件开发 , C++ ,