HDU - 6582 Path (最短路+最小割)

题意：给定一个n个点m条边的有向图，每条边有个长度，可以花费等同于其长度的代价将其破坏掉，求最小的花费使得从1到n的最短路变长。

解法：先用dijkstra求出以1为源点的最短路，并建立最短路图(只保留d[v]=d[u]+e[i].c的边(u,v))，跑个最大流即可。

Dinic:

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll N=1e4+,inf=0x3f3f3f3f3f3f3f3fll;
 ll n,m;
 struct MF {
     ll hd[N],ne,d[N],g[N],cur[N];
     struct E {ll v,cp,nxt;} e[N<<];
     void init() {memset(hd,-,sizeof hd),ne=;}
     void addedge(ll u,ll v,ll cp) {
         e[ne]= {v,cp,hd[u]},hd[u]=ne++;
         e[ne]= {u,,hd[v]},hd[v]=ne++;
     }
     int bfs(ll s,ll t) {
         queue<ll> q;
         memset(d,-,sizeof d);
         d[s]=,q.push(s);
         while(q.size()) {
             ll u=q.front();
             q.pop();
             for(ll i=hd[u]; ~i; i=e[i].nxt)if(e[i].cp) {
                     ll v=e[i].v;
                     if(!~d[v])d[v]=d[u]+,q.push(v);
                 }
         }
         return ~d[t];
     }
     ll dfs(ll t,ll u,ll f) {
         if(u==t||!f)return f;
         ll ret=;
         for(ll& i=cur[u]; ~i; i=e[i].nxt) {
             ll v=e[i].v;
             if(d[v]==d[u]+) {
                 ll df=dfs(t,v,min(f,e[i].cp));
                 e[i].cp-=df,e[i^].cp+=df,f-=df,ret+=df;
                 if(!f)return ret;
             }
         }
         return ret;
     }
     ll dinic(ll s,ll t) {
         ll ret=;
         while(bfs(s,t)) {
             for(ll i=; i<=n; ++i)cur[i]=hd[i];
             ret+=dfs(t,s,inf);
         }
         return ret;
     }
 } mf;
 struct SP {
     ll hd[N],ne,dp[N];
     struct E {ll v,c,nxt;} e[N];
     void init() {memset(hd,-,sizeof hd),ne=;}
     void addedge(ll u,ll v,ll c) {e[ne]= {v,c,hd[u]},hd[u]=ne++;}
     struct D {
         ll u,g;
         bool operator<(const D& b)const {return b.g>g;}
     };
     priority_queue<D> q;
     void upd(ll u,ll ad) {if(dp[u]>ad)dp[u]=ad,q.push({u,ad});}
     void dij(ll s) {
         memset(dp,inf,sizeof dp);
         upd(s,);
         while(q.size()) {
             ll u=q.top().u,g=q.top().g;
             q.pop();
             if(dp[u]!=g)continue;
             for(ll i=hd[u]; ~i; i=e[i].nxt) {
                 ll v=e[i].v,c=e[i].c;
                 upd(v,g+c);
             }
         }
     }
     void solve() {
         dij();
         for(ll u=; u<=n; ++u) {
             for(ll i=hd[u]; ~i; i=e[i].nxt) {
                 ll v=e[i].v,c=e[i].c;
                 if(dp[v]==dp[u]+c)mf.addedge(u,v,c);
             }
         }
         printf("%lld\n",mf.dinic(,n));
     }
 } sp;
 int main() {
     ll T;
     for(scanf("%lld",&T); T--;) {
         sp.init(),mf.init();
         scanf("%lld%lld",&n,&m);
         while(m--) {
             ll u,v,c;
             scanf("%lld%lld%lld",&u,&v,&c);
             sp.addedge(u,v,c);
         }
         sp.solve();
     }
     return ;
 }

ISAP:

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll N=1e4+,inf=0x3f3f3f3f3f3f3f3fll;
 ll n,m;
 struct MF {
     ll hd[N],ne,d[N],g[N],cur[N];
     struct E {ll v,cp,nxt;} e[N<<];
     void init() {memset(hd,-,sizeof hd),ne=;}
     void addedge(ll u,ll v,ll cp) {
         e[ne]= {v,cp,hd[u]},hd[u]=ne++;
         e[ne]= {u,,hd[v]},hd[v]=ne++;
     }
     void bfs(ll s) {
         queue<ll> q;
         memset(d,-,sizeof d);
         memset(g,,sizeof g);
         ++g[d[s]=],q.push(s);
         while(q.size()) {
             ll u=q.front();
             q.pop();
             for(ll i=hd[u]; ~i; i=e[i].nxt) {
                 ll v=e[i].v;
                 if(!~d[v])++g[d[v]=d[u]+],q.push(v);
             }
         }
     }
     ll dfs(ll s,ll t,ll u,ll f) {
         if(u==t||!f)return f;
         ll ret=;
         for(ll& i=cur[u]; ~i; i=e[i].nxt) {
             ll v=e[i].v;
             if(d[v]==d[u]-) {
                 ll df=dfs(s,t,v,min(f,e[i].cp));
                 e[i].cp-=df,e[i^].cp+=df,f-=df,ret+=df;
                 if(!f)return ret;
             }
         }
         if(!--g[d[u]])d[s]=n+;
         ++g[++d[u]],cur[u]=hd[u];
         return ret;
     }
     ll isap(ll s,ll t) {
         ll ret=;
         bfs(t);
         for(ll i=; i<=n; ++i)cur[i]=hd[i];
         while(d[s]<n+)ret+=dfs(s,t,s,inf);
         return ret;
     }
 } mf;
 struct SP {
     ll hd[N],ne,dp[N];
     struct E {ll v,c,nxt;} e[N];
     void init() {memset(hd,-,sizeof hd),ne=;}
     void addedge(ll u,ll v,ll c) {e[ne]= {v,c,hd[u]},hd[u]=ne++;}
     struct D {
         ll u,g;
         bool operator<(const D& b)const {return b.g>g;}
     };
     priority_queue<D> q;
     void upd(ll u,ll ad) {if(dp[u]>ad)dp[u]=ad,q.push({u,ad});}
     void dij(ll s) {
         memset(dp,inf,sizeof dp);
         upd(s,);
         while(q.size()) {
             ll u=q.top().u,g=q.top().g;
             q.pop();
             if(dp[u]!=g)continue;
             for(ll i=hd[u]; ~i; i=e[i].nxt) {
                 ll v=e[i].v,c=e[i].c;
                 upd(v,g+c);
             }
         }
     }
     void solve() {
         dij();
         for(ll u=; u<=n; ++u) {
             for(ll i=hd[u]; ~i; i=e[i].nxt) {
                 ll v=e[i].v,c=e[i].c;
                 if(dp[v]==dp[u]+c)mf.addedge(u,v,c);
             }
         }
         printf("%lld\n",mf.isap(,n));
     }
 } sp;
 int main() {
     ll T;
     for(scanf("%lld",&T); T--;) {
         sp.init(),mf.init();
         scanf("%lld%lld",&n,&m);
         while(m--) {
             ll u,v,c;
             scanf("%lld%lld%lld",&u,&v,&c);
             sp.addedge(u,v,c);
         }
         sp.solve();
     }
     return ;
 }

这道题Dinic居然比ISAP快，我的代码ISAP跑了156ms，而Dinic只跑了93ms..