POJ-3268.SilverCowParty.(最短路 + 图的转置)

　　本题思路：对原图和原图的逆图分别用一次最短路，找出最大值即可。

一开始是我是对每个顶点spfa搜了一波，结果判题时间巨长，还好这个题的数据量不是很大，所以就用了另一种思路。

　　参考代码：spfa单结点爆搜版

 #include <iostream>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <algorithm>
 using namespace std;

 const int maxn =  + , INF = 0x3f3f3f3f;
 struct edge {
     int to, cost;
 };
 int n, m, x, ans, dist[maxn];
 vector<edge> G[maxn];
 bool vis[maxn];

 void addedge(int u, int v, int w) {
     G[u].push_back({v, w});
 }

 int spfa(int source, int aid) {
     memset(vis, false, sizeof vis);
     for(int i = ; i <= n; i ++)
         dist[i] = (i == source ?  : INF);
     queue <int> Q;
     Q.push(source);
     vis[source] = true;
     while(!Q.empty()) {
         int now = Q.front();
         Q.pop();
         for(int i = ; i < G[now].size(); i ++) {
             edge e = G[now][i];
             if(dist[e.to] > dist[now] + e.cost) {
                 dist[e.to] = dist[now] + e.cost;
                 if(!vis[e.to]) Q.push(e.to);
             }
         }
     }
     return dist[aid];
 }

 int main () {
     ans = ;
     int u, v, w;
     cin >> n >> m >> x;
     for(int i = ; i <= m; i ++) {
         cin >> u >> v >> w;
         addedge(u, v, w);
     }
     for(int i = ; i <= n; i ++)
         ans = max(spfa(i, x) + spfa(x, i), ans);
     cout << ans << endl;
     return ;
 }

　　对原图和逆图分别进行一次Dijkstra的终极版本

 #include <iostream>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <algorithm>
 using namespace std;

 const int maxn =  + , INF = 0x3f3f3f3f;
 struct edge {
     int to, cost;
 };
 int n, m, x, k, ans, dist1[maxn], dist2[maxn], G[maxn][maxn];;
 bool vis[maxn];

 void Dijkstra(int lowcost[]) {
     for(int i = ; i <= n; i ++)
         lowcost[i] = INF;
     lowcost[x] = ;
     memset(vis, false, sizeof vis);
     for(int i = ; i <= n; i ++) {
         int minf = INF;
         for(int j = ; j <= n; j ++)
             if(minf > lowcost[j] && !vis[j]) {
                 k = j;
                 minf = lowcost[j];
             }
         vis[k] = true;
         for(int j = ; j <= n; j ++)
             if(!vis[j] && lowcost[j] > lowcost[k] + G[k][j])
                 lowcost[j] = lowcost[k] + G[k][j];
     }
 }

 int main () {
     ans = ;
     int u, v, w;
     cin >> n >> m >> x;
     for(int i = ; i <= n; i ++)
         for(int j = ; j <= n; j ++)
             if(i == j) G[i][j] = ;
             else G[i][j] = INF;
     for(int i = ; i <= m; i ++) {
         cin >> u >> v >> w;
         G[u][v] = min(G[u][v], w);
     }
     Dijkstra(dist1);
     for(int i = ; i <= n; i ++)
         for(int j = ; j < i; j ++)
             if(G[i][j]) swap(G[i][j], G[j][i]);
     Dijkstra(dist2);
     for(int i = ; i <= n; i ++)
         ans = max(ans, dist1[i] + dist2[i]);
     cout << ans << endl;
     return ;
 }