對於所有資料 n≤1000，m≤5000，每條道路的花費≤10000

分析

假如一開始就是偶環必然要刪掉。
假如是奇環，相交之後會形成一個偶環，只要邊不相交就行。
考慮到兒子數很少，設狀態 $f_{i,S}$

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 5050;
int n,m,ans;
int final[N],nex[N],to[N],tot;
int f[
 
N][N],g[N],fmx[N],cnt[N];

void link(int x,int y) {
	to[++tot] = y,nex[tot] = final[x],final[x] = tot;
}

int e[N][3],ec,rk[N],fa[N],dep[N];

struct edge{
	int x,y,v;
}; vector<edge> up[N];

void dfs(int x) {
	int r=0; dep[x]=dep[fa[x]]+1;
	for (int i = final[x]; i; i = nex[i]) {
		int y = to[i]
 
;
		if (y != fa[x]) {
			rk[y] = r++;
			fa[y] = x;
			dfs(y);
		}
	}
	cnt[x] = r;
}

int lca(int x,int y) {
	if (dep[x]<dep[y]) swap(x,y);
	while (dep[x]>dep[y]) x=fa[x];
	if (x==y) return x;
	while (x!=y) x=fa[x],y=fa[y];
	return x;
}

int getsum(int u,int tu,int &near) {
	int ret = fmx[u];
	for (; u != tu; u = fa[u]) {
		ret += g[u];
		if (fa[u] == tu) near |= 1<<rk[u];
	}
	return ret;
}

edge a;
int count(int x) {
	int ret = 0; while (x) x-=x&-x,ret++;
	return ret;
}
void solve(int x) {
	for (int i = final[x]; i; i=nex[i]) {
		int y = to[i]; if (y != fa[x]) {
			solve(y);
		}
	}
	
	while (up[x].size()) {
		a = up[x].back(); up[x].pop_back();
		int s = 0;
		int v = getsum(a.x,x,s);
		v += getsum(a.y,x,s) + a.v;
		for (int i = 0; i < (1<<cnt[x]); i++) if ((i & s)==0) {
			f[x][i | s] = max(f[x][i | s], f[x][i] + v);
		}
	}
	
	for (int i = 0; i < (1<<cnt[x]); i++) {
		for (int j = final[x]; j; j=nex[j]) {
			int y = to[j]; if (y != fa[x]) {
				if ((i & (1<<rk[y]))==0) {
					f[x][i] += fmx[y];
				}
			}
		}
		for (int j = final[x]; j; j=nex[j]) {
			int y = to[j]; if (y != fa[x]) {
				if ((i & (1<<rk[y]))==0) {
					g[y] = max(g[y], f[x][i] - fmx[y]);
				}
			}
		}
		fmx[x]=max(fmx[x],f[x][i]);
	}
}

int he;
int main() {
	freopen("zujijihua.in","r",stdin);
	freopen("zujijihua.out","w",stdout);
	cin>>n>>m;
	for (int i = 1; i <= m; i++) {
		int u,v,c;scanf("%d %d %d",&u,&v,&c);
		if (c==0)
			link(u,v),link(v,u);
		else {
			e[++ec][0] = u,e[ec][1] = v,e[ec][2] = c;
		}
		ans+=c;
	}

	dfs(1);
	for (int i = 1; i <= ec; i++) {
		int l = lca(e[i][0],e[i][1]);
		he += e[i][2];
		if ((dep[e[i][0]] + dep[e[i][1]] - dep[l] * 2 + 1) & 1) {
			up[l].push_back((edge){e[i][0],e[i][1],e[i][2]});
		}
	}
	solve(1);
	cout<<he-fmx[1]<<endl;
}