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\n",dinic.Maxflow(1,n));
    }
}