題目鏈接

題意

N個城市M條路徑，給定起點A，終點B，求有幾條從A到B的最短路（其中每經過的路徑不能重復）

解題思路

先用最短路求出A到B的最短路Min,也求出A到每個城市的距離dis[N],然後反向求B到A的最短路，得到B到每個城市的最短距離dis2[N]，然後遍歷每條路徑edge，如果dis[edge.from] + edge.len + dis2[edge.to]== Min,就說明這條路徑一定是A到B的最短路中會經過的路徑，，每條路徑的容量為1，把每條符合條件的路徑加入到最大流的圖中，建完圖後，以A為源點，B為匯點跑最大流即可（用EdmondsKarp會超時，跑Dinic即可）

AC代碼

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 1e5+5;
const int INF = 0x3f3f3f3f;
struct Edge 
{
    int from,to,cap,flow;
    Edge(){}
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};

struct Dinic
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void init(int n){
        for(int i=0;i<n;i++) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BFS()
    {
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;
        while(!Q.empty()){
            int x = Q.front();
            Q.pop();
            for(int i=0;i<G[x].size();i++){
                Edge& e = edges[G[x][i]];
                if(!vis[e.to]&&e.cap > e.flow){
                    vis[e.to] = 1;
                    d[e.to] = d[x]+1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x,int a){
        if(x==t || a==0)return a;
        int flow = 0,f;
        for(int &i=cur[x];i<G[x].size();i++){
            Edge& e = edges[G[x][i]];
            if(d[x]+1==d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0){
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if(a==0)break;
            }
        }
        return flow;
    }

    int Maxflow(int s,int t){
        this->s = s,this->t = t;
        int flow = 0;
        while(BFS()){
            memset(cur,0,sizeof(cur));
            flow += DFS(s,INF);
        }
        return flow;
    }
    
};

vector<Edge> G1[maxn];
vector<Edge> G2[maxn];
int N,M;
int dis[maxn],dis2[maxn];
bool vis[maxn];

void dijkstra(int A,vector<Edge> G[maxn],int dis[maxn])
{
    memset(vis,0,sizeof(vis));
    priority_queue<pii,vector<pii>,greater<pii> > Q;
    dis[A] = 0;
    Q.push(pii(dis[A],A));
    while(!Q.empty()){
        pii t = Q.top();
        Q.pop();
        int d = t.first;
        int u = t.second;
        for(int i=0;i<G[u].size();i++){
            Edge e = G[u][i];
            if(e.cap + d < dis[e.to]){
                dis[e.to] = e.cap + d;
                Q.push(pii(dis[e.to],e.to));
            }
        }
    }
}

int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T = 0;
    cin >> T;
    Dinic ek;
    while(T--){
        cin >> N >> M;
        ek.init(N);
        for(int i=0;i<=N;i++){
            G1[i].clear();
            G2[i].clear();
        }
        int a,b,c;
        Edge e;
        for(int i=0;i<M;i++){
            cin >> a >> b >> c;
            e.from = a;
            e.to = b;
            e.cap = c;
            G1[e.from].push_back(e);
            swap(e.to,e.from);
            G2[e.from].push_back(e);
        }
        int A,B;
        cin >> A >> B;
        memset(dis,0x3f,sizeof(dis));
        dijkstra(A,G1,dis);             //正向跑最短路
        memset(dis2,0x3f,sizeof(dis2));
        int Min = dis[B];               
        dijkstra(B,G2,dis2);            //反向跑最短路
        for(int i=1;i<=N;i++){
            for(int j=0;j<G1[i].size();j++){
                e = G1[i][j];
                if(dis[e.from] + e.cap + dis2[e.to]==Min){
                    ek.AddEdge(e.from,e.to,1);
                }
            }
        }
        cout << ek.Maxflow(A,B) << endl;
    }
    
    return 0;
}
 

hdu3416Marriage Match IV (網絡流+最短路)