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hdu 1853 Cyclic Tour && hdu 3435 A new Graph Game（简单KM算法）

2024-07-14 10:39:50   224人阅读

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1478    Accepted Submission(s): 750


Problem Description
There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?
 

Input
There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).
 

Output
Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1. 
 

Sample Input
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
 

Sample Output
42
-1


Hint
 In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.
 

这是1853代码，后一题建成双向边就行了，代码类似。

#include"stdio.h"
#include"string.h"
#define N 105
#define mmax(a,b) ((a)>(b)?(a):(b))
#define mmin(a,b) ((a)<(b)?(a):(b))
const int inf=(int)1e8;
int g[N][N],slack[N],n;
int link[N],lx[N],ly[N];
int visx[N],visy[N];
int find(int k)
{
    int i;
    visx[k]=1;
    for(i=1;i<=n;i++)
    {
        if(visy[i])
            continue;
        int d=lx[k]+ly[i]-g[k][i];
        if(d==0)
        {
            visy[i]=1;
            if(link[i]==-1||find(link[i]))
            {
                link[i]=k;
                return 1;
            }
        }
        else
            slack[i]=mmin(slack[i],d);
    }
    return 0;
}
int KM()
{
    int i,j;
    memset(ly,0,sizeof(ly));
    memset(link,-1,sizeof(link));
    for(i=1;i<=n;i++)
    {
        lx[i]=-inf;
        for(j=1;j<=n;j++)
        {
            lx[i]=mmax(lx[i],g[i][j]);
        }
    }
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            slack[j]=inf;
        }
        while(1)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(find(i))
                break;
            else
            {
                int d=inf;
                for(j=1;j<=n;j++)
                {
                    if(!visy[j])
                        d=mmin(slack[j],d);
                }
                for(j=1;j<=n;j++)
                {
                    if(visx[j])
                        lx[j]-=d;
                    if(visy[j])
                        ly[j]+=d;
                }
            }
        }
    }
    int ans=0;
    for(i=1;i<=n;i++)
    {
        if(link[i]==-1||g[link[i]][i]==-inf)
            break;
        ans+=g[link[i]][i];
    }
    if(i<=n)
        return -1;
    return -ans;
}
int main()
{
    int i,j,u,v,w,m;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        for(i=1;i<=n;i++)        //注意初始化边为最小值
        {
            for(j=1;j<=n;j++)
            {
                g[i][j]=-inf;
            }
        }
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&w);
            g[u][v]=mmax(g[u][v],-w);
        }
        printf("%d\n",KM());
    }
    return 0;
}


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