标签：des style blog class code int
题意：题目给出一个有向图 ， 找若干个圈，使得每个结点切好属于一个圈，并且所有圈的总长度最小 ， 如果没有满足条件的就输出 ‘N‘ 。这题可以学到：以后求有向图中的环 ， 可以用二分图来求。
代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
using namespace std;
#define maxn 120
#define INF 0xffffff
int grap[maxn][maxn];
int n;
int cx[maxn] , cy[maxn];
int lx[maxn] , ly[maxn];
int pre[maxn];
int s[maxn] , t[maxn];
int slack[maxn];
void init()
{
memset(grap , -1 , sizeof(grap));
memset(cx , -1 , sizeof(cx));
memset(cy , -1 , sizeof(cy));
}
int findpath(int u)
{
s[u] = 1;
for(int i = 1; i <= n; i++)
{
if(grap[u][i] != -1)
{
if(lx[u]+ly[i]==grap[u][i] && !t[i])
{
t[i] = 1;
if(cy[i] == -1 || findpath(cy[i]))
{
cy[i] = u;
cx[u] = grap[u][i];
return 1;
}
}
else slack[i] = min(slack[i] , grap[u][i]-lx[u]-ly[i]);
}
}
return 0;
}
int match()
{
int i , j;
for(i = 1; i <= n; i++)
{
ly[i] = 0;
lx[i] = INF;
for(j = 1; j <= n; j++)
if(grap[i][j] != -1)
lx[i] = min(lx[i] , grap[i][j]);
}
for(i = 1; i <= n; i++)
{
for(; ;)
{
for(j = 1; j <= n; j++)
{
s[j]=t[j] = 0;
slack[j] = INF;
}
if(findpath(i)) break;
int a = INF;
for(j = 1; j <= n; j++)
if(!t[j]) a = min(a , slack[j]);
if(a == INF) return -1;
for(j = 1; j <= n; j++)
{
if(s[j]) lx[j] += a;
if(t[j]) ly[j] -= a;
}
}
}
return 1;
}
int main()
{
while(scanf("%d" , &n) && n)
{
init();
int i , j , x , y;
for(i = 1; i <= n; i++)
{
for(; ;)
{
scanf("%d" , &x);
if(x == 0) break;
scanf("%d" , &y);
if(grap[i][x] == -1 || grap[i][x] > y)
grap[i][x] = y;
}
}
x = match();
if(x == -1)
{
cout<<"N"<<endl;
}
else
{
x = 0;
for(i = 1; i <= n; i++)
x += cx[i];
cout<<x<<endl;
}
}
return 0;
}LA 3353 Optimal Bus Route Design 二分匹配和有向图中的环,布布扣,bubuko.com
LA 3353 Optimal Bus Route Design 二分匹配和有向图中的环
标签：des style blog class code int
原文地址：http://blog.csdn.net/zengchen__acmer/article/details/25281105
