题意:给定一个有向图,让你找出若干个图,使得每个点恰好属于一个圈,并且总的权和最小。

析:每个点都有唯一的一个圈,也就是说每一点都有唯一的后继,那么我们就可以转换成求一个图的最小权的最佳完全匹配,可以用最小费用流来求,

先把每个结点拆成两个点,假设是x,y,然后建立一个源点,向每个点的x连一条容量为1的边,建立一个汇点,每个点的y向汇点连一条容量为1的边,

每条边u,v,也连接一条容量为1,费用为权值的边,最小求一个最小费用流即可。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 2000 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int from, to, cap, flow;
LL cost;
}; struct MCMF{
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];
LL d[maxn];
int p[maxn];
int a[maxn]; void init(int n){
this->n = n;
for(int i = 0; i < n; ++i) G[i].clear();
edges.clear();
} void addEdge(int from, int to, int cap, LL cost){
edges.push_back((Edge){from, to, cap, 0, cost});
edges.push_back((Edge){to, from, 0, 0, -cost});
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
} bool BellmanFord(int s, int t, int &flow, LL &cost){
for(int i = 0; i < n; ++i) d[i] = INF;
memset(inq, 0, sizeof inq);
d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF; queue<int> q;
q.push(s);
while(!q.empty()){
int u = q.front(); q.pop();
inq[u] = 0;
for(int i = 0; i < G[u].size(); ++i){
Edge &e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost){
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap-e.flow);
if(!inq[e.to]) q.push(e.to), inq[e.to] = 1;
}
}
}
if(d[t] == INF) return false;
flow += a[t];
cost += d[t] * a[t];
int u = t;
while(u != s){
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
u = edges[p[u]].from;
}
return true;
} int minCost(int s, int t, LL &cost){
int flow = 0; cost = 0;
while(BellmanFord(s, t, flow, cost));
return flow;
}
};
MCMF mcmf; int main(){
while(scanf("%d", &n) == 1 && n){
mcmf.init(n+n+2);
int v; LL val;
int s = 0, t = n+n+1;
for(int i = 1; i <= n; ++i){
// mcmf.addEdge(i, i+n, 1, 0);
mcmf.addEdge(s, i, 1, 0);
mcmf.addEdge(i+n, t, 1, 0);
while(scanf("%d", &v) == 1 && v){
scanf("%lld", &val);
mcmf.addEdge(i, v+n, 1, val);
}
} LL ans;
if(mcmf.minCost(s, t, ans) == n) printf("%lld\n", ans);
else printf("N\n");
}
return 0;
}

最新文章

  1. 微软开源代码编辑器monaco-editor
  2. [LeetCode 118] - 杨辉三角形(Pascal&#39;s Triangle)
  3. (转)ios限制控制器旋转
  4. @dynamic 与 @synthesize
  5. java构建学生管理系统(一)
  6. 珍爱生命,远离JS=&gt;JS避坑记
  7. Linux命令top 详解
  8. java算法之超级丑数
  9. C\C++控制台程序隐藏方法总结
  10. SQL 字段修改
  11. 后台登录(包含验证码)的php代码实现
  12. BZOJ1975 [Sdoi2010]魔法猪学院 k短路
  13. codeforces 578a//A Problem about Polyline// Codeforces Round #320 (Div. 1)
  14. max_result_window
  15. How to change windows applicatioin&#39;s position via Win32 API
  16. 2018 Multi-University Training Contest 6-oval-and-rectangle(hdu 6362)-题解
  17. 代替Reflection(反射)的一些方法(转)
  18. PHP加密解密数字,适用于URL加密。
  19. Qt基于tcp协议网络编程
  20. QToolBox学习笔记

热门文章

  1. ECharts整合HT&amp;#160;for&amp;#160;Web的网络拓扑图应用
  2. 利用CH341A编程器刷新BIOS,恢复BIOS,妈妈再也不用担心BIOS刷坏了
  3. WWDC2014苹果的“软件”发布会
  4. Java RESTful 框架
  5. 微信小程序 如何使用globalData
  6. iOS 7 中 StoryBoard 总体缩放
  7. 关于mongodb副本集读写分离 及 日志切换
  8. Html.DropDownListFor的选项值为字符型问题
  9. photoswipe 实现图片的单击放大
  10. shapes