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 = 1000 + 5, 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 = 1; i <= n; i ++)
        dist[i] = (i == source ? 0 : INF);
    queue <int> Q;
    Q.push(source);
    vis[source] = true;
    while(!Q.empty()) {
        int now = Q.front();
        Q.pop();
        for(int i = 0; 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 = 0;
    int u, v, w;
    cin >> n >> m >> x;
    for(int i = 1; i <= m; i ++) {
        cin >> u >> v >> w;
        addedge(u, v, w);
    }
    for(int i = 1; i <= n; i ++)
        ans = max(spfa(i, x) + spfa(x, i), ans);
    cout << ans << endl;
    return 0;
}

 　　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 = 1000 + 5, 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 = 1; i <= n; i ++)
        lowcost[i] = INF;
    lowcost[x] = 0;
    memset(vis, false, sizeof vis);
    for(int i = 1; i <= n; i ++) {
        int minf = INF;
        for(int j = 1; j <= n; j ++) 
            if(minf > lowcost[j] && !vis[j]) {
                k = j;
                minf = lowcost[j];
            }
        vis[k] = true;
        for(int j = 1; j <= n; j ++)
            if(!vis[j] && lowcost[j] > lowcost[k] + G[k][j])
                lowcost[j] = lowcost[k] + G[k][j];
    }
}

int main () {
    ans = 0;
    int u, v, w;
    cin >> n >> m >> x;
    for(int i = 1; i <= n; i ++)    
        for(int j = 1; j <= n; j ++)
            if(i == j) G[i][j] = 0;
            else G[i][j] = INF;
    for(int i = 1; i <= m; i ++) {
        cin >> u >> v >> w;
        G[u][v] = min(G[u][v], w);
    }
    Dijkstra(dist1);
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j < i; j ++)
            if(G[i][j]) swap(G[i][j], G[j][i]);
    Dijkstra(dist2);
    for(int i = 1; i <= n; i ++)
        ans = max(ans, dist1[i] + dist2[i]);
    cout << ans << endl;
    return 0;
}