题意及思路:https://blog.csdn.net/mengxiang000000/article/details/52472696

代码:

#define Hello the_cruel_world!
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<utility>
#include<cmath>
#include<climits>
#include<deque>
#include<functional>
#include<numeric>
#define max(x,y) ((x) > (y) ? (x) : (y))
#define min(x,y) ((x) < (y) ? (x) : (y))
#define lowbit(x) ((x) & (-(x)))
#define FRIN freopen("C:\\Users\\Administrator.MACHENI-KA32LTP\\Desktop\\1.in", "r", stdin)
#define FROUT freopen("C:\\Users\\Administrator.MACHENI-KA32LTP\\Desktop\\1.out", "w", stdout)
#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define outd(x) printf("%d\n", x)
#define outld(x) printf("%lld\n", x)
#define memset0(arr) memset(arr, 0, sizeof(arr))
#define il inline
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int maxn = 110;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const double Pi = acos(-1.0);
il int read_int() {
char c;
int ret = 0, sgn = 1;
do { c = getchar(); } while ((c < '0' || c > '9') && c != '-');
if (c == '-') sgn = -1; else ret = c - '0';
while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
return sgn * ret;
}
il ll read_ll() {
char c;
ll ret = 0, sgn = 1;
do { c = getchar(); } while ((c < '0' || c > '9') && c != '-');
if (c == '-') sgn = -1; else ret = c - '0';
while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
return sgn * ret;
}
il ll quick_pow(ll base, ll index) {
ll res = 1;
while (index) {
if (index & 1)res = res * base % mod;
base = base * base % mod;
index >>= 1;
}
return res;
}
struct edge {
int to, capacity, cost, rev;
edge() {}
edge(int to, int _capacity, int _cost, int _rev) :to(to), capacity(_capacity), cost(_cost), rev(_rev) {}
};
struct Min_Cost_Max_Flow {
int V, H[maxn + 5], dis[maxn + 5], PreV[maxn + 5], PreE[maxn + 5];
vector<edge> G[maxn + 5];
//调用前初始化
void Init(int n) {
V = n;
for (int i = 0; i <= V; ++i)G[i].clear();
}
//加边
void Add_Edge(int from, int to, int cap, int cost) {
G[from].push_back(edge(to, cap, cost, G[to].size()));
G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
}
//flow是自己传进去的变量,就是最后的最大流,返回的是最小费用
int Min_cost_max_flow(int s, int t, int f, int& flow, int limit) {
int res = 0; fill(H, H + 1 + V, 0);
while (f) {
priority_queue <pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>> > q;
fill(dis, dis + 1 + V, INF);
dis[s] = 0; q.push(pair<int, int>(0, s));
while (!q.empty()) {
pair<int, int> now = q.top(); q.pop();
int v = now.second;
if (dis[v] < now.first)continue;
for (int i = 0; i < G[v].size(); ++i) {
edge& e = G[v][i];
if (e.capacity > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to]) {
dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
PreV[e.to] = v;
PreE[e.to] = i;
q.push(pair<int, int>(dis[e.to], e.to));
}
}
}
if (dis[t] == INF)break;
for (int i = 0; i <= V; ++i)H[i] += dis[i];
int d = f;
for (int v = t; v != s; v = PreV[v])
d = min(d, G[PreV[v]][PreE[v]].capacity);
if(H[t] != 0) {
d = min(d, limit / H[t]);
}
if(d == 0) break;
// printf("d: %d\n", d);
// printf("H[t]: %d\n", H[t]);
// printf("flow :%d\n", flow);
limit -= d * H[t];
int tmp = d * H[t];
// printf("tmp: %d\n", tmp);
f -= d; flow += d; res += d*H[t];
// printf("flow :%d\n", flow);
for (int v = t; v != s; v = PreV[v]) {
edge& e = G[PreV[v]][PreE[v]];
e.capacity -= d;
G[v][e.rev].capacity += d;
}
}
return res;
}
int Max_cost_max_flow(int s, int t, int f, int& flow, int limit) {
int res = 0;
fill(H, H + 1 + V, 0);
while (f) {
priority_queue <pair<int, int>> q;
fill(dis, dis + 1 + V, -INF);
dis[s] = 0;
q.push(pair<int, int>(0, s));
while (!q.empty()) {
pair<int, int> now = q.top(); q.pop();
int v = now.second;
if (dis[v] > now.first)continue;
for (int i = 0; i < G[v].size(); ++i) {
edge& e = G[v][i];
if (e.capacity > 0 && dis[e.to] < dis[v] + e.cost + H[v] - H[e.to]) {
dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
PreV[e.to] = v;
PreE[e.to] = i;
q.push(pair<int, int>(dis[e.to], e.to));
}
}
}
if (dis[t] == -INF)break;
for (int i = 0; i <= V; ++i)H[i] += dis[i];
int d = f;
for (int v = t; v != s; v = PreV[v])d = min(d, G[PreV[v]][PreE[v]].capacity);
d = min(limit, d);
f -= d; flow += d;
limit -= d;
res += d*H[t];
for (int v = t; v != s; v = PreV[v]) {
edge& e = G[PreV[v]][PreE[v]];
e.capacity -= d;
G[v][e.rev].capacity += d;
}
if(limit == 0) return res;
}
return res;
}
};
int n, k, s, t, flow, m;
Min_Cost_Max_Flow MCMF;
int main()
{
int x, y, z;
scanf("%d%d", &n, &k);
s = 1, t = n;
MCMF.Init(t);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; j++) {
scanf("%d", &z);
if(z != 0) {
MCMF.Add_Edge(i, j, z, 0);
MCMF.Add_Edge(i, j, k, 1);
}
}
}
MCMF.Min_cost_max_flow(s, t, INF, flow, k);
printf("%d\n", flow);
return 0;
}

  

Codeforces 362E 费用流的更多相关文章

  1. Codeforces 730I [费用流]

    /* 不要低头,不要放弃,不要气馁,不要慌张 题意: 给两行n个数,要求从第一行选取a个数,第二行选取b个数使得这些数加起来和最大. 限制条件是第一行选取了某个数的条件下,第二行不能选取对应位置的数. ...

  2. Codeforces 708D 费用流 (呃我们的考试题)

    NB的题目背景 输入输出一样 考试的时候貌似只有gzz一个人搞出来了 %gzz 思路: 分情况讨论 add(x,y,C,E) C是费用 E是流量 1. f>c add(x,y,2,inf),ad ...

  3. Codeforces 362E Petya and Pipes 费用流建图

    题意: 给一个网络中某些边增加容量,增加的总和最大为K,使得最大流最大. 费用流:在某条边增加单位流量的费用. 那么就可以2个点之间建2条边,第一条给定边(u,v,x,0)这条边费用为0 同时另一条边 ...

  4. CodeForces 164C Machine Programming 费用流

    Machine Programming 题目连接: http://codeforces.com/problemset/problem/164/B Descriptionww.co One remark ...

  5. Codeforces Gym 100002 E "Evacuation Plan" 费用流

    "Evacuation Plan" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...

  6. Codeforces Gym 101190M Mole Tunnels - 费用流

    题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...

  7. Codeforces 280D k-Maximum Subsequence Sum [模拟费用流,线段树]

    洛谷 Codeforces bzoj1,bzoj2 这可真是一道n倍经验题呢-- 思路 我首先想到了DP,然后矩阵,然后线段树,然后T飞-- 搜了题解之后发现是模拟费用流. 直接维护选k个子段时的最优 ...

  8. BZOJ 3836 Codeforces 280D k-Maximum Subsequence Sum (模拟费用流、线段树)

    题目链接 (BZOJ) https://www.lydsy.com/JudgeOnline/problem.php?id=3836 (Codeforces) http://codeforces.com ...

  9. Codeforces 708D 上下界费用流

    给你一个网络流的图 图中可能会有流量不平衡和流量>容量的情况存在 每调整一单位的流量/容量 需要一个单位的花费 问最少需要多少花费使得原图调整为正确(可行)的网络流 设当前边信息为(u,v,f, ...

随机推荐

  1. 【LeetCode】随机化算法 random(共6题)

    [384]Shuffle an Array(2019年3月12日) Shuffle a set of numbers without duplicates. 实现一个类,里面有两个 api,struc ...

  2. springboot全局字符编码设置

    1.在application.properties中设置 #编码格式 spring.http.encoding.force=true spring.http.encoding.charset=UTF- ...

  3. ansible笔记(三)--模块讲解

    ansible 常用命令 ansible-doc ansible-playbook ansible-vault ansible-console ansible-galaxy ansible-pull ...

  4. JSON 简单例子

    代码: json [ { "title" : "a", "num" : 1 }, { "title" : "b ...

  5. 第01章 Spring概述

    第01章 Spring概述 1.Spring概述 ①Spring是一个开源框架 ②Spring为简化企业级开发而生,使用Spring,JavaBean就可以实现很多以前要靠EJB才能实现的功能.同样的 ...

  6. 数据库与缓存:2.Redis数据库的基本知识

    1.属于什么类型的数据库 not only sql  非关系型数据库,与传统的关系型数据库不同,存储形式都是kv形式. 2.特点 几乎不支持事务,key-value形式存储,支持队列和缓存(可以设置数 ...

  7. JAVA学习笔记--初探hash与map

    先推荐一篇文章 http://blog.csdn.net/cownew/article/details/6478993 也为自己mark

  8. uwsgi部署django项目

    一.更新系统软件包 yum update -y 二.安装软件管理包及依赖 yum -y groupinstall "Development tools" yum install o ...

  9. Python3 使用HTMLTestRunner.py 报错ImportError: No module named 'StringIO'处理方法

    HTMLTestRunner.py文件是基于Python2的语法,python3使用需要修改语法: 1.>>> 94行 import io # import StringIO 2.& ...

  10. list的过滤操作

    假设 l = ['abc', 'mn', 'aq', 'liuming'] 我要过滤出以a开头的元素,方法有以下两种 方法1: l = ['abc', 'mn', 'aq', 'liuming'] l ...