• 2019 Multi-University Training Contest 1 Path(最短路+最小割)


    题意:给你n个点 m条边 现在你能够堵住一些路 问怎样能让花费最少且让1~n走的路比最短路的长度要长

    思路:先跑一边最短路 建一个最短路图 然后我们跑一边最大流求一下最小割即可

    #include <bits/stdc++.h>
    using namespace std;
    const double pi = acos(-1.0);
    const int maxn = 1e4+7;
    const int inf = 0x3f3f3f3f;
    const double eps = 1e-6;
    typedef long long ll;
    const ll mod = 1e9+7;
    struct edge{
        int next,to; ll w;
    };
    edge e[maxn<<1];
    int head[maxn],cnt;
    int vis[maxn];
    ll d[maxn];
    void init(){
        cnt=0;
        memset(head,0,sizeof(head));
        memset(vis,0,sizeof(vis));
        memset(d,inf,sizeof(d));
    }
    void add(int u, int v, int w){
        e[++cnt]={head[u],v,w};
        head[u] = cnt;
    }
    void dij(int s){
        priority_queue<pair<ll,int> > q;
        d[s]=0;
        q.push(make_pair(0,s));
        while(!q.empty()){
            int u=q.top().second;
            q.pop();
            if(vis[u]) continue;
            vis[u]=1;
            for(int i=head[u];i;i=e[i].next){
                int v=e[i].to; int w=e[i].w;
                if(d[v]>d[u]+w){
                    d[v]=d[u]+w;
                    q.push(make_pair(-d[v],v));
                }
            }
        }
    }
    struct Edge {
      ll from, to, cap, flow;
      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];
      int d[maxn], cur[maxn];
      bool vis[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];
      }
    
      ll DFS(int x, ll a) {
        if (x == t || a == 0) return a;
        ll 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;
      }
    
      ll Maxflow(int s, int t) {
        this->s = s;
        this->t = t;
        ll flow = 0;
        while (BFS()) {
          memset(cur, 0, sizeof(cur));
          flow += DFS(s, inf);
        }
        return flow;
      }
    } dinic;
    int bian[maxn][3];
    int main(){
        int t;
        scanf("%d",&t);
        while(t--){
            init();
            int n,m;
            scanf("%d%d",&n,&m);
            for(int i=1;i<=m;i++){
                int x,y,c; scanf("%d%d%d",&x,&y,&c);
                bian[i][0]=x; bian[i][1]=y; bian[i][2]=c;
                add(x,y,c);
            }
            dij(1);
            dinic.init(n);
            for(int i=1;i<=m;i++){
                if(d[bian[i][1]]==d[bian[i][0]]+bian[i][2]){
                    dinic.AddEdge(bian[i][0],bian[i][1],bian[i][2]);
                }
            }
            printf("%lld
    ",dinic.Maxflow(1,n));
        }
    } 
    View Code
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  • 原文地址:https://www.cnblogs.com/wmj6/p/11348247.html
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