题意:
给定一棵有 n 个节点的树,每条边有边权 w(u,v),用最少的路径覆盖所有的边,使得每条边被覆盖的次数等于其边权。
思路:
最坏情况是每条边都单独覆盖w次。逐个合并。
#include <cstdio>
#include <queue>
#include <vector>
#include <cstring>
#include <algorithm>
#define fi first
#define se second
#define pii pair<int,int>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long LL;
const int maxn = 1000000+5;
int n, deg[maxn], maxDeg[maxn]; // n,度数,最大度数
int main()
{
freopen("in.txt","r",stdin);
int T; scanf("%d",&T);
int kase = 1;
while(T--){
scanf("%d",&n);
for(int i = 0; i <= n; ++i) deg[i] = maxDeg[i] = 0;
int ans = 0;
for(int i = 0; i < n-1; ++i){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
deg[u] += w; deg[v] += w;
maxDeg[u] = max(maxDeg[u], w);
maxDeg[v] = max(maxDeg[v], w);
ans += w;
}
for(int i = 1; i <= n; ++i){
if(maxDeg[i]*2 <= deg[i]) ans-= deg[i]/2;
else ans-= deg[i] - maxDeg[i];
}
printf("Case #%d: %d\n", kase++, ans);
}
return 0;
}