2018 Benelux Algorithm Programming Contest (BAPC 18)I In Case of an Invasion, Please. . .

题意:一副无向有权图,每个点有一些人,某些点是避难所(有容量),所有人要去避难所,问最小时间所有人都到达避难所,

题解:dij+二分+最大流check,注意到避难所最多10个,先挨个dij求到避难所的时间,然后二分时间,在这个时间之内的建边,s向避难所建边,流量是避难所容量,可达的避难所向点建边,流量inf,点向t建边,流量为人的个数,看能不能满流即可,wa点:maxflow里的inf忘改,maxn开小了= =

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-9;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+100,maxn=4000000+10,inf=0x3f3f3f3f;

vector<pil>v[N];
ll d[12][N],a[N],b[20];
priority_queue<pli,vector<pli>,greater<pli>>q;
void dij(int s,int id)
{
    memset(d[id],INF,sizeof d[id]);
    d[id][s]=0;
    q.push(mp(d[id][s],s));
    while(!q.empty())
    {
        pli u=q.top();q.pop();
        if(u.fi>d[id][u.se])continue;
        for(pil x:v[u.se])
        {
            if(d[id][u.se]+x.se<d[id][x.fi])
            {
                d[id][x.fi]=d[id][u.se]+x.se;
                q.push(mp(d[id][x.fi],x.fi));
            }
        }
    }
}
struct edge{
    int to,Next;
    ll c;
}e[maxn];
int s,t,cnt,head[N],cur[N];
void init(){cnt=0;memset(head,-1,sizeof head);}
void add(int u,int v,ll c)
{
    e[cnt].to=v;e[cnt].c=c;e[cnt].Next=head[u];head[u]=cnt++;
    e[cnt].to=u;e[cnt].c=0;e[cnt].Next=head[v];head[v]=cnt++;
}
int dis[N];
bool bfs()
{
    queue<int>q;
    memset(dis,-1,sizeof dis);
    dis[s]=1;q.push(s);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=head[x];~i;i=e[i].Next)
        {
            int y=e[i].to;
            if(dis[y]==-1&&e[i].c>0)
            {
                dis[y]=dis[x]+1;
                q.push(y);
            }
        }
    }
    return dis[t]!=-1;
}
ll dfs(int u,ll mx)
{
    if(u==t)return mx;
    ll flow=0,f;
    for(int &i=cur[u];~i;i=e[i].Next)
    {
        int x=e[i].to;
        if(dis[x]==dis[u]+1&&e[i].c>0&&(f=dfs(x,min(mx-flow,1ll*e[i].c))))
        {
            e[i].c-=f;
            e[i^1].c+=f;
            flow+=f;
            if(flow==mx)break;
        }
    }
    if(flow==0)dis[u]=-2;
    return flow;
}
ll maxflow()
{
    ll ans=0,f;
    while(bfs())
    {
        for(int i=0;i<=t;i++)cur[i]=head[i];
        while((f=dfs(s,INF)))ans+=f;
    }
    return ans;
}
int n,m,c;
bool ok(ll x)
{
    init();s=n+c+1,t=n+c+2;
    ll sum=0;
    for(int i=1;i<=c;i++)
    {
        add(s,i,b[i]);
        for(int j=1;j<=n;j++)if(d[i][j]<=x)add(i,c+j,INF);
    }
    for(int i=1;i<=n;i++)add(c+i,t,a[i]),sum+=a[i];
    return maxflow()==sum;
}
int main()
{
    scanf("%d%d%d",&n,&m,&c);
    for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
    for(int i=1;i<=m;i++)
    {
        int a,b;ll c;scanf("%d%d%lld",&a,&b,&c);
        v[a].pb(mp(b,c));v[b].pb(mp(a,c));
    }
    vector<ll>te;
    for(int i=1,x;i<=c;i++)
    {
        scanf("%d%lld",&x,&b[i]),dij(x,i);
        for(int j=1;j<=n;j++)te.pb(d[i][j]);
    }
    sort(te.begin(),te.end());te.erase(unique(te.begin(),te.end()),te.end());
    int l=-1,r=te.size();
    while(l<r-1)
    {
        int mid=(l+r)>>1;
        if(ok(te[mid]))r=mid;
        else l=mid;
    }
    printf("%lld\n",te[r]);
    return 0;
}
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