标签:匹配 while eps lld [] design signed div 完全
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;
}
UVa 1349 Optimal Bus Route Design (最佳完美匹配)
标签:匹配 while eps lld [] design signed div 完全
原文地址:http://www.cnblogs.com/dwtfukgv/p/6416068.html