2019杭电多校第一场 E - Path（最短路+最小割）

题目链接：E - Path HDU - 6582

题意：给定n个点，m条有向边，首先求出最短路，然后在这张新图上跑最小割

 

最短路和最小割都是板子，我们现在关注的就是如何构造出这张图：

for(int u=1;u<=n;++u)
    for(int i=head[u];i!=-1;i=edge[i].next)
        if(dis[edge[i].v]=dis[u]+edge[i].w){
            ++id;
            line[id].u=u;
            line[id].v=v;
            line[id].w=w;
        }

ok。

#include
using namespace std;

typedef long long ll;

const int maxn=1e4+7;
const int maxm=2e4+7;

struct Edge{
    int v,next;
    ll w;
}edge[maxm],edge1[maxm];

int head[maxn],head1[maxn],top,top1;

void init(int n){
    top=0;
    for(int i=1;i<=n;++i) head[i]=-1;
}

void init1(int n){
    top1=0;
    for(int i=1;i<=n;++i) head1[i]=-1;
}

void add(int u,int v,ll w){
    edge[top].v=v;
    edge[top].w=w;
    edge[top].next=head[u];
    head[u]=top++;
}

void add1(int u,int v,ll w){
    edge1[top1].v=v;
    edge1[top1].w=w;
    edge1[top1].next=head1[u];
    head1[u]=top1++;
}

priority_queue > yq;
const ll inf=0x3f3f3f3f3f3f3f3f;

ll dis[maxn];
bool vis[maxn];
void dij(int n){
    while(!yq.empty()) yq.pop();
    for(int i=1;i<=n;++i) vis[i]=0,dis[i]=inf;
    int u,v;
    ll w;
    dis[1]=0;
    yq.push({0,1});
    while(!yq.empty()){
        u=yq.top().second; yq.pop();
        if(vis[u]) continue;
        vis[u]=1;
        for(int i=head[u];i!=-1;i=edge[i].next){
            v=edge[i].v;
            w=edge[i].w;
            if(dis[v]>dis[u]+w){
                dis[v]=dis[u]+w;
                yq.push({-dis[v],v});
            }
        }
    }
}

queue q;
int s,t;
ll maxflow;
int d[maxn];
bool bfs(){
    memset(d,0,sizeof(d));
    while(!q.empty()) q.pop();
    q.push(s);
    int v;
    ll w;
    d[s]=1;
    while(!q.empty()){
        int u=q.front(); q.pop();
        for(int i=head1[u];i!=-1;i=edge1[i].next){
            v=edge1[i].v;
            w=edge1[i].w;
            if(w&&!d[v]){
                q.push(v);
                d[v]=d[u]+1;
                if(v==t) return 1;
            }
        }
    }
    return 0;
}

ll dinic(int u,ll flow){
    if(u==t) return flow;
    ll res=flow,k;
    for(int i=head1[u];(i!=-1)&&res;i=edge1[i].next){
        ll w=edge1[i].w;
        int v=edge1[i].v;
        if(w&&d[v]==d[u]+1){
            k=dinic(v,min(res,w));
            if(!k) d[v]=0;
            edge1[i].w-=k;
            edge1[i^1].w+=k;
            res-=k;
        }
    }
    return flow-res;
}

int main(){
    int T,n,m;
    int u,v,w;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        init(n);
        while(m--){
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,1LL*w);
        }
        dij(n);
        if(dis[n]==inf) printf("0\n");
        else{
            init1(n);
            int hh=0;
            for(u=1;u<=n;++u)
                for(int i=head[u];i!=-1;i=edge[i].next)
                    if(dis[edge[i].v]==dis[u]+edge[i].w){
                        //cout<