EOJ3745. 迷宮【二分答案+最大流】

題目鏈接

---

  這裏的點和邊的個數都是比較小的，當然，不是說可以暴力了。可以用分層圖的方式來解決這題，就是，我們給每個點每個時間對應的流，假設ans時間可以完成最大流，於是就可以利用最大流再加上每個點到下一個點的時間來進行計算了，如果最大流等於人數了，說明這個時間是可行時間，不然就是時間偏小了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e4 + 7, maxM = 3e5 + 7;
int people, Point, M;
int head[maxN], cnt;
struct Eddge
{
    int nex, to; ll flow;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
} edge[maxM];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, 0); }
struct Max_Flow
{
    int gap[maxN], d[maxN], que[maxN], cur[maxN], ql, qr, S, T, node;
    inline void init()
    {
        for(int i=0; i<=node + 1; i++)
        {
            gap[i] = d[i] = 0;
            cur[i] = head[i];
        }
        ++gap[d[T] = 1];
        que[ql = qr = 1] = T;
        while(ql <= qr)
        {
            int x = que[ql ++];
            for(int i=head[x], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(!d[v]) { ++gap[d[v] = d[x] + 1]; que[++qr] = v; }
            }
        }
    }
    inline ll aug(int x, ll FLOW)
    {
        if(x == T) return FLOW;
        int flow = 0;
        for(int &i=cur[x], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(d[x] == d[v] + 1)
            {
                ll tmp = aug(v, min(FLOW, edge[i].flow));
                flow += tmp; FLOW -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
                if(!FLOW) return flow;
            }
        }
        if(!(--gap[d[x]])) d[S] = node + 1;
        ++gap[++d[x]]; cur[x] = head[x];
        return flow;
    }
    inline ll max_flow()
    {
        init();
        ll ret = aug(S, INF);
        while(d[S] <= node) ret += aug(S, INF);
        return ret;
    }
} mf;
struct Graph
{
    int u, v; ll w;
    void In() { scanf("%d%d%lld", &u, &v, &w); }
} Save[maxM];
void init()
{
    cnt = 0;
    for(int i=0; i<=mf.node; i++) head[i] = -1;
}
int S, T;
bool check(int lim)
{
    mf.node = 1 + Point * (lim + 1);
    mf.T = mf.node;
    init();
    for(int i=1; i<=Point; i++)
    {
        for(int j=1; j<=lim; j++)
        {
            _add(i + (j - 1) * Point, i + j * Point, INF);
        }
    }
    _add(mf.S, S, people);
    for(int i=1; i<=lim; i++) _add(T + i * Point, mf.T, INF);
    for(int i=1, u, v; i<=M; i++)
    {
        u = Save[i].u; v = Save[i].v;
        ll w = Save[i].w;
        for(int j=1; j<=lim; j++)
        {
            _add(u + (j - 1) * Point, v + j * Point, w);
        }
    }
    return mf.max_flow() == people;
}
int main()
{
    scanf("%d%d%d%d%d", &people, &Point, &M, &mf.S, &mf.T);
    mf.node = Point; T = mf.T;
    for(int i=1; i<=M; i++) Save[i].In();
    int l = 1, r = people + Point, mid, ans = 0;
    S = 0; swap(mf.S, S);
    while(l <= r)
    {
        mid = HalF;
        if(check(mid))
        {
            ans = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}