BZOJ 3931 Dijkstra+网络流

思路：
（我能说按照题意模拟么）

用long long
inf 要开大…….

//By SiriusRen
#include 
#include 
#include 
#include 
using namespace std;
#define int long long
#define N 1017
#define M 999999
#define inf 0x3f3f3f3f3f3f3f3fll
int n,m,xx[M],yy[M],zz[M],first[N],next[M],v[M],w[M],tot,dis[N],vis[N],tmp,ans;
struct Node{int wei,now;}jy;
bool operator < (Node a,Node b){return a.wei>b.wei;}
void Add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}
struct Dijkstra{
    void add(int x,int y,int z){Add(x,y,z),Add(y,x,z);}
    void solve(){
        memset(dis,0x3f,sizeof(dis)),dis[1]=0;
        priority_queuepq;jy.now=1,pq.push(jy);
        while(!pq.empty()){
            Node t=pq.top();pq.pop();
            if(vis[t.now])continue;
            vis[t.now]=1;
            for(int i=first[t.now];~i;i=next[i])
                if(dis[v[i]]>dis[t.now]+w[i])
                    dis[v[i]]=dis[t.now]+w[i],
                    jy.now=v[i],jy.wei=dis[v[i]],pq.push(jy);
        }
    }
}dijk;
struct Dinic{
    void add(int x,int y,int z){Add(x,y,z),Add(y,x,0);}
    bool tell(){
        memset(vis,-1,sizeof(vis)),vis[1]=0;
        queue<int>q;q.push(1);
        while(!q.empty()){
            int t=q.front();q.pop();
            for(int i=first[t];~i;i=next[i])
                if(!~vis[v[i]]&&w[i])
                    vis[v[i]]=vis[t]+1,q.push(v[i]);
        }return ~vis[2*n];
    }
    int zeng(int x,int y){
        if(x==2*n)return y;
        int r=0;
        for(int i=first[x];~i;i=next[i])
            if(vis[v[i]]==vis[x]+1&&w[i]){
                int t=zeng(v[i],min(y-r,w[i]));
                w[i]-=t,w[i^1]+=t,r+=t;
            }
        if(!r)vis[x]=-1;
        return r;
    }
    void flow(){
        while(tell())while(tmp=zeng(1,inf))ans+=tmp;
        printf("%lld\n",ans);
    }
}dinic;
void init(){memset(first,-1,sizeof(first)),tot=0;}
signed main(){
    init(),scanf("%lld%lld",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%lld%lld%lld",&xx[i],&yy[i],&zz[i]);
        dijk.add(xx[i],yy[i],zz[i]);
    }
    dijk.solve(),init();
    for(int i=1;i<=m;i++){
        if(dis[yy[i]]==dis[xx[i]]+zz[i])dinic.add(xx[i]+n,yy[i],inf);
        if(dis[xx[i]]==dis[yy[i]]+zz[i])dinic.add(yy[i]+n,xx[i],inf);
    }
    for(int i=1;i<=n;i++){
        scanf("%lld",&tmp);
        if(i!=1&&i!=n)dinic.add(i,i+n,tmp);
        else dinic.add(i,i+n,inf);
    }
    dinic.flow();
}

转载于:https://www.cnblogs.com/SiriusRen/p/6532113.html