meaning of the title

Given an undirected connected graph, each edge has a deletion cost. Ask how much it costs to delete edges so that there is just one simple path between 1 and n.
n<=15

Analysis

Just because we can't do it well, we'll consider how many sides we can add at most.
My approach is to squeeze out each path and find that if there is only one simple path, no two points on the path can be in the same point of the double connected component. Then we can connect a connected block to a point every time. Then enumerate which connection is fast connected to which point each time, so that the complexity will be one more n, but it is not necessary, as long as the transfer of connected blocks is added in the transitional path.

Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
#define MAX(x,y) x=max(x,y)
using namespace std;

const int N=18;
const int M=32775;

int n,m,bin[N],num[M],ma[N][M],sum[M],f[N][M],g[M];

int read()
{
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

int lowbit(int x)
{
    return x&(-x);
}

int main()
{
    n=read();m=read();
    bin[0]=1;
    for (int i=1;i<=n;i++) bin[i]=bin[i-1]*2,num[bin[i]]=i;
    int tot=0;
    for (int i=1;i<=m;i++)
    {
        int x=read()-1,y=read()-1,z=read();tot+=z;
        ma[x][bin[y]]=ma[y][bin[x]]=z;
    }
    for (int i=0;i<n;i++)
        for (int j=1;j<bin[n];j++)
            ma[i][j]=ma[i][j-lowbit(j)]+ma[i][lowbit(j)];
    for (int i=1;i<bin[n];i++) sum[i]=sum[i-lowbit(i)]+ma[num[lowbit(i)]][i-lowbit(i)];
    for (int i=0;i<n;i++)
        for (int j=0;j<bin[n];j++)
            f[i][j]=-inf;
    f[0][1]=0;
    for (int j=0;j<bin[n];j++)
        for (int i=0;i<n-1;i++)
        {
            if (f[i][j]<0) continue;
            for (int k=0;k<n;k++)
                if (!(j&bin[k])&&ma[i][bin[k]]) MAX(f[k][j+bin[k]],f[i][j]+ma[i][bin[k]]);
        }
    for (int i=0;i<bin[n];i++) g[i]=f[n-1][i];
    for (int i=1;i<bin[n];i++)
    {
        if (g[i]<0) continue;
        int r=i^(bin[n]-1);
        for (int s=r;s;s=(s-1)&r)
        {
            int mxv=0;
            for (int j=i;j;j-=lowbit(j)) mxv=max(mxv,sum[s+lowbit(j)]);
            MAX(g[i+s],g[i]+mxv);
        }
    }
    printf("%d",tot-g[bin[n]-1]);
    return 0;
}