6 9 1 2 5 2 3 5 3 1 10 3 4 12 4 1 8 4 6 11 5 4 7 5 6 9 6 5 4 6 5 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1
42 -1HintIn the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
/*
KM算法 O(nx*nx*ny)
求最大权匹配(最佳匹配)
若求最小权匹配,可将权值取相反数,结果取相反数
点的标号从0开始
*/
const int N=110;
int nx,ny; //两边的点数
int g[N][N]; //二分图描述,g赋初值为-INF
int linker[N],lx[N],ly[N]; //y 中各点匹配状态,x,y中的点的标号
int slack[N];
bool visx[N],visy[N];
bool flag;
bool DFS(int x)
{
visx[x]=true;
for (int y=0;y<ny;y++)
{
if (visy[y]) continue;
int tmp=lx[x]+ly[y]-g[x][y];
if (tmp==0)
{
visy[y]=true;
if (linker[y]==-1||DFS(linker[y]))
{
linker[y]=x;
return true;
}
}
else if (slack[y]>tmp)
slack[y]=tmp;
}
return false;
}
int KM()
{
flag=true;
memset(linker,-1,sizeof(linker));
memset(ly,0,sizeof(ly));
for (int i=0;i<nx;i++) //赋初值,lx置为最大值
{
lx[i]=-INF;
for (int j=0;j<ny;j++)
{
if (g[i][j]>lx[i])
lx[i]=g[i][j];
}
}
for (int x=0;x<nx;x++)
{
for (int i=0;i<ny;i++)
slack[i]=INF;
while (true)
{
memset(visx,false,sizeof(visx));
memset(visy,false,sizeof(visy));
if (DFS(x)) break;
int d=INF;
for (int i=0;i<ny;i++)
if (!visy[i]&&d>slack[i])
d=slack[i];
for (int i=0;i<nx;i++)
if (visx[i])
lx[i]-=d;
for (int i=0;i<ny;i++)
{
if (visy[i])
ly[i]+=d;
else
slack[i]-=d;
}
}
}
int res=0;
for (int i=0;i<ny;i++)
{
if (linker[i]==-1||g[linker[i]][i]<=-INF) //有的点不能匹配的话return-1
{
flag=false;
break;
}
res+=g[linker[i]][i];
}
return res;
}
int n,m;
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
int i,j;
while (~sff(n,m))
{
for (i=0;i<=n;i++)
for (j=0;j<=n;j++)
g[i][j]=-INF;
nx=ny=n;
int u,v,w;
for (i=0;i<m;i++)
{
sfff(u,v,w);
u--;
v--;
if (-w>g[u][v])
g[u][v]=-w;
}
int ans=KM();
ans=-ans;
if (!flag)
pf("-1\n");
else
pf("%d\n",ans);
}
return 0;
}
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Cyclic Tour (hdu 1853 二分图最小权问题)
原文:http://blog.csdn.net/u014422052/article/details/46761341