问题描述

总时间限制: 2000ms内存限制: 65536kB

描述

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

输入

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

输出

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

样例输入

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
         (0)5 (1)2 (3)2 (4)1 (5)4

样例输出

15
6

提示

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.


http://162.105.81.202/ps2007/docs/IAP-5%20maxflow1.pdf

问题解决

/*
 * Dinic algo for max flow
 *
 * This implementation assumes that #nodes, #edges, and capacity on each edge <= INT_MAX,
 * which means INT_MAX is the best approximation of INF on edge capacity.
 * The total amount of max flow computed can be up to LLONG_MAX (not defined in this file),
 * but each 'dfs' call in 'dinic' can return <= INT_MAX flow value.
 */
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <limits.h>
#include <string.h>
#include <assert.h>
#include <queue>
#include <vector>
#include <algorithm>

#define N (200)
#define M (80800)

typedef long long LL;

using namespace std;

struct edge {
  int v, cap, next;
};
edge e[M];

int head[N], level[N], cur[N];
int num_of_edges;

/*
 * When there are multiple test sets, you need to re-initialize before each
 */
void dinic_init(void) {
  num_of_edges = 0;
  memset(head, -1, sizeof(head));
  return;
}

int add_edge(int u, int v, int c1, int c2) {
  int& i=num_of_edges;

  assert(c1>=0 && c2>=0 && c1+c2>=0); // check for possibility of overflow
  e[i].v = v;
  e[i].cap = c1;
  e[i].next = head[u];
  head[u] = i++;

  e[i].v = u;
  e[i].cap = c2;
  e[i].next = head[v];
  head[v] = i++;
  return i;
}

void print_graph(int n) {
  for (int u=0; u<n; u++) {
    printf("%d: ", u);
    for (int i=head[u]; i>=0; i=e[i].next) {
      printf("%d(%d)", e[i].v, e[i].cap);
    }
    printf("\n");
  }
  return;
}

/*
 * Find all augmentation paths in the current level graph
 * This is the recursive version
 */
int dfs(int u, int t, int bn) {
  if (u == t) return bn;
  int left = bn;
  for (int &i=cur[u]; i>=0; i=e[i].next) {
    int v = e[i].v;
    int c = e[i].cap;
    if (c > 0 && level[u]+1 == level[v]) {
      int flow = dfs(v, t, min(left, c));
      if (flow > 0) {
        e[i].cap -= flow;
        e[i^1].cap += flow;
        cur[u] = i;
        left -= flow;
        if (!left) break;
      }
    }
  }
  if (left > 0) level[u] = 0;
  return bn - left;
}

bool bfs(int s, int t) {
  memset(level, 0, sizeof(level));
  level[s] = 1;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (u == t) return true;
    for (int i=head[u]; i>=0; i=e[i].next) {
      int v = e[i].v;
      if (!level[v] && e[i].cap > 0) {
        level[v] = level[u]+1;
        q.push(v);
      }
    }
  }
  return false;
}

LL dinic(int s, int t) {
  LL max_flow = 0;

  while (bfs(s, t)) {
    memcpy(cur, head, sizeof(head));
    max_flow += dfs(s, t, INT_MAX);
  }
  return max_flow;
}

int upstream(int s, int n) {
  int cnt = 0;
  vector<bool> visited(n);
  queue<int> q;
  visited[s] = true;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i=head[u]; i>=0; i=e[i].next) {
      int v = e[i].v;
      if (e[i].cap > 0 && !visited[v]) {
        visited[v] = true;
        q.push(v);
        cnt++;
      }
    }
  }
  return cnt; // excluding s
}

int main(){
    int n,np,nc,m;
    while(cin>>n>>np>>nc>>m){
        dinic_init();
        int s=n,t=n+1;
        for(int i=0;i<m;i++){
            int from,to,val;
            char tmp;
            cin>>tmp>>from>>tmp>>to>>tmp>>val;
            add_edge(from,to,val,0);
        }
        for(int i=0;i<np;i++){
            int node,val;
            char tmp;
            cin>>tmp>>node>>tmp>>val;
            add_edge(s,node,val,0);
        }
        for(int i=0;i<nc;i++){
            int node,val;
            char tmp;
            cin>>tmp>>node>>tmp>>val;
            add_edge(node,t,val,0);
        }
        printf("%lld\n",dinic(s,t));
    }
    return 0;
}

标签：head,return,power,level,int,cap,461Power,POJ,dinic
来源： https://blog.csdn.net/qq_39921575/article/details/112790856