JZOJ4716. The Flying Birds __ the puzzle

Original question: NOI2012 Food Festival topic purport

There are N-n-type birds, there are m-m holes, the first I-type birds have pi p_i only, through the J-J hole Need T (i,j) T (i,j) time. Each bird waits for a bird to pass through the same hole before it passes. Ask for the minimum total wait time.

Data Constraint
N≤40,m≤100,∑pi≤800,t (i,j) ≤1000 n\leq40,m\leq100,\sum p_i\leq800,t (i,j) \leq1000

For a Hole J J, the last passing bird I I to the answer of the contribution for T (I,J) T (I,J), the penultimate contribution for 2∗t (I,J) 2*t (i,j), the rest etc.
Consider the cost stream, which (Flow,cost) (Flow,cost) represents an edge in which the flow costs are cost: s S to each type of bird Company (pi,0) (p_i,0) each hole into a ∑pi \sum p_i point, each species of birds first these Point Edge (1,k∗t (I,J)) (1,k*t (i,j)) per hole to T-T Company (1,0) (1,0)

The number of sides is more, so use dynamic add edge. SRC

#include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <

Algorithm> #include <queue> using namespace std;

#define N + #define M 100000 + typedef long LL;
BOOL Flag[n][10*n], vis[m];
int node[2*m], next[2*m], c[2*m], cost[2*m], head[m], tot = 1;

int d[10*m], dist[m], pre[m], rec[2*m], bel[2*m];
int tim[n][n], p[n];
int n, m, S, T, sump;

ll ans; void link (int u, int v, int w, int fee) {Node[++tot] = V, next[tot] = Head[u], C[tot] = W, cost[tot] = fee,
    Head[u] = tot;
Node[++tot] = u, Next[tot] = Head[v], C[tot] = 0, Cost[tot] =-fee, head[v] = tot;
    BOOL Spfa () {memset (Vis, 0, sizeof (VIS));
    memset (Dist, sizeof (Dist));
    int i = 0, j = 1;
    D[1] = S;
    Dist[s] = 0;
    Vis[s] = 1;
        while (I < j) {i + +;
        int now = D[i]; for (int p = Head[now]; p. p = next[p]) {if (! C[P]) continue;
            if (Dist[now] + cost[p] < Dist[node[p]]) {dist[node[p]] = Dist[now] + cost[p];
                PRE[NODE[P]] = p;
                    if (!vis[node[p]]) {vis[node[p]] = 1;
                    D[++J] = node[p];
                if (dist[d[i+1]] > Dist[d[j]) swap (d[i+1], d[j]);
    }} Vis[now] = 0; }