2020牛客多校训练1 H Minimumcost Flow(费用流)

题目链接：https://ac.nowcoder.com/acm/contest/5666/H

题意：给你一个n个点m条边的图，q次询问每次给出u和v，每条边的容量都为u/v， 初始流量为1，询问满流时候的最小费用。

首先将每条边的容量设为1跑一遍费用流 。那么对于某次询问容量变为u/v，只要将起始流量设为v，每条边容量设为u之后得到的答案再除回v即可。

而如何实现上述操作，只要先预处理出容量为1时每次求增广路所需要的花费，然后优先选花费最小的v/u(向上整除)条增广路，如果v%u!=0,

那么最后一条增广路上的容量应该是v%u而不是u。这样子q次询问的答案，全部可以通过一开始容量为1的费用流得出的结果线性表示。具体实现见代码。

#include<bits/stdc++.h>
#define ll long long
#define PB push_back
#define endl '\n'
#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ull unsigned long long
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define lowbit(x) (x & (-x))
#define rep(i, a, b) for(int i = a ; i <= b ; ++ i)
#define per(i, a, b) for(int i = b ; i >= a ; -- i)
#define clr(a, b) memset(a, b, sizeof(a))
#define in insert
#define random(x) (rand()%x)
#define PII(x, y) make_pair(x, y)
#define fi first
#define se second
#define pi acos(-1)
using namespace std;
const int maxn = 2e6 + 1000;
const ll mod = 998244353;
const int M = 100 + 10;
const int N = 50 + 10;
int n, m, s, t, cnt;
int head[N], vis[N], inq[N], cur[N];
ll dis[N], sum[M<<1];
ll maxflow, cost;
vector<ll> path;
struct node{
    int to, next;
    ll cost, flow;
}edge[M<<1];
void add(int u, int v, int cost, int flow){
    edge[cnt].to = v;
    edge[cnt].cost = cost;
    edge[cnt].flow = flow;
    edge[cnt].next = head[u];
    head[u] = cnt ++;
}
bool spfa(){
    rep(i, 1, n) dis[i] = 1e18, inq[i] = 0, cur[i] = head[i];
    queue<int> que; dis[s] = 0;
    inq[s] = 1; que.push(s);
    while(!que.empty()){
        int u = que.front();
        que.pop();
        inq[u] = 0;
        for(int i = head[u] ; ~ i ; i = edge[i].next){
            int v = edge[i].to;
            if(dis[v] > dis[u] + edge[i].cost && edge[i].flow){
                dis[v] = dis[u] + edge[i].cost;
                if(!inq[v]){
                    inq[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    return dis[t] == 1e18 ? 0 : 1;
}
ll dfs(int u, ll flow){
    if(u == t){
        vis[t] = 1;
        maxflow += flow;
        return flow;
    }
    ll used = 0;
    vis[u] = 1;
    for(int i = cur[u] ; ~ i ; i = edge[i].next){
        int v = edge[i].to;
        cur[u] = i;
        if((v == t || !vis[v]) && edge[i].flow && dis[v] == dis[u] + edge[i].cost){
            ll minflow = dfs(v, min(flow - used, edge[i].flow));
            if(minflow){
                cost += minflow * edge[i].cost;
                edge[i].flow -= minflow;
                edge[i^1].flow += minflow;
                used += minflow;
            }
            if(used == flow) break;
        }
    }
    return used;
}
void dinic(){
    while(spfa()){
        vis[t] = 1;
        path.PB(dis[t]);
        while(vis[t]){
            clr(vis, 0);
            dfs(s, 1e18);
        }
    }
}
signed main(){
    ll u, v, w; int q;
    while(~scanf("%d %d", &n, &m)){
        clr(head, -1); clr(vis, 0);
        path.clear(); maxflow = cost = 0;
        s = 1; t = n; cnt = 0;
        while(m --){
            scanf("%lld %lld %lld", &u, &v, &w);
            add(u, v, w, 1);
            add(v, u, -w, 0);
        }
        scanf("%d", &q);
        dinic();
        ll sz = path.size();
        rep(i, 0, sz-1) sum[i+1] = sum[i] + path[i];
        while(q --){
            scanf("%lld %lld", &u, &v);
            if(sz * u < v) {puts("NaN"); continue;}
            ll x = v / u, y = v % u;
            ll ans = sum[x] * u + path[x] * y;
            ll tmp = __gcd(ans, v);
            cout << ans / tmp << '/' << v / tmp << endl;
        }
    }
    return 0;
}