HDU - 5889 Barricade 最短路+网络流

http://acm.hdu.edu.cn/showproblem.php?pid=5889

题意：敌人在n点，敌人会走n到1的最短路径进攻你了，现在你需要在路径上放置障碍，求能阻挡所有敌人放置最少价值的障碍。

思路：我们先对原图跑最短路，然后重新建图跑dicnic 注意使用弧优化

#include<bits/stdc++.h>
#define maxn 1005
using namespace std;
#define ll long long
const int INF = 0x3f3f3f3f;
int head2[1005],To2[maxn*20],Next2[maxn*20],dis[maxn],cost2[maxn*20],vis[maxn];
int head[1005],To[maxn*20],Next[maxn*20];
int flow[maxn*20],dep[maxn],cur[maxn];
int cnt,S,T;
void add(int u,int v,int w){
    Next2[cnt]=head2[u];
    head2[u]=cnt;
    To2[cnt]=v;
    cost2[cnt++]=w;
}
void add2(int u,int v,int w){
    Next[cnt]=head[u];
    head[u]=cnt;
    To[cnt]=v;
    flow[cnt++]=w;
}
void init(){
    memset(head,-1,sizeof(head));
    memset(head2,-1,sizeof(head2));
    memset(dis,INF,sizeof(dis)); 
    memset(vis,0,sizeof(vis)); 
    memset(dep,0,sizeof(dep));
    memset(cur,0,sizeof(cur));
    cnt=0;
}
void dij(){
    priority_queue<pair<int,int> , vector<pair<int,int> > ,greater<pair<int,int> > > q;
    q.push({0,1});
    dis[1]=0;
    while(!q.empty()){
        int u=q.top().second;
        q.pop();
        if(vis[u]) continue;
        vis[u]=1;
        for(int i=head2[u];i!=-1;i=Next2[i]){
            int v=To2[i];
            if(dis[v]>dis[u]+1){
                dis[v]=dis[u]+1;
                q.push({dis[v],v});
            }
        }
    }
}
int dfs(int u,int cost){
    if(u==T) return cost;
    for(int &i=cur[u];i!=-1;i=Next[i]){
        if(dep[To[i]]==dep[u]+1&&flow[i]>0){
            int dis=dfs(To[i],min(flow[i],cost));
            if(dis>0){
                flow[i]-=dis;
                flow[i^1]+=dis;
                return dis;
            }
        }
    }
    return 0;
}
int bfs(){
    queue<int> q;
    q.push(1);
    memset(dep,0,sizeof(dep));
    dep[1]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=Next[i]){
            if(flow[i]>0&&dep[To[i]]==0){
                dep[To[i]]=dep[u]+1;
                q.push(To[i]);
            }
        }
    }
    if(dep[T]>0) return 1;
    return 0;
}
int dicnic(){
    int ans=0;
    while(bfs()){
        int cost;
        for(int i=1;i<=T+2;i++){
            cur[i]=head[i];
        }
        while(cost=dfs(1,INF))
            ans+=cost;
    }
    return ans;
}
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 u,v,w;
            scanf("%d %d %d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w); 
        }
        dij();
    //    cout<<dis[n]<<endl;
        cnt=0;
        T=n;
        for(int i=1;i<=n;i++){
            for(int j=head2[i];j!=-1;j=Next2[j]){
            //    cout>>
                if(dis[i]+1==dis[To2[j]]){
                    add2(i,To2[j],cost2[j]);
                    add2(To2[j],i,0);
                //    cout<<i<<"  "<<To[j]<<endl;
                }
            }
        }
        cout<<dicnic()<<"\n"; 
    }
    return 0;
}