题目大意:
给一张无向图,可以将图中权值为v的边修改为k-v,求修改后的最小生成树边权和。
题目分析:
最小生成树的环切定理
:任何非树边一定比对应树链上的所有边权值要大。否则我们可以将链上最大的树边删去而连接这一条边。
运用这个性质,先求出原图的最小生成树,然后再来枚举边,如果枚举到一条边:
- 它是最小生成树中的边,如果更优, 那么更新答案。
- 它不是最小生成树中的边,找到它连接的两点在最小生成树中的链上边权的最大值,如果山出发这条边,加入枚举的这条更优,就更新答案
code
#include<bits/stdc++.h>
using namespace std;
const int N = 1050, M = 1e6 + 50, OO = 0x3f3f3f3f;
int n;
int m, k, ans1, ans2;
struct node{
int x, y, t;
inline bool operator < (const node &b) const{
return t < b.t;
}
}edge[M];
struct node2{
int ecnt, adj[N], go[N << 1], nxt[N << 1], len[N << 1], used[M], d[N][N], anc[N];
node2(){}
inline void init(){
ecnt = 0;
memset(adj, 0, sizeof adj);
memset(used, 0, sizeof used);
for(int i = 1; i <= n; i++) anc[i] = i;
}
inline int getAnc(int x){
return x == anc[x] ? x : (anc[x] = getAnc(anc[x]));
}
inline void addEdge(int u, int v, int t){
nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v, len[ecnt] = t;
}
inline int kruskals(){
int ret = 0;
sort(edge + 1, edge + m + 1);
for(int i = 1; i <= m; i++){
int fx = getAnc(edge[i].x), fy = getAnc(edge[i].y);
if(fx != fy){
anc[fx] = fy;
ret += edge[i].t;
used[i] = 1;
addEdge(edge[i].x, edge[i].y, edge[i].t);
addEdge(edge[i].y, edge[i].x, edge[i].t);
}
}
return ret;
}
inline void dfs(int now, int u, int f, int mx){
d[now][u] = d[u][now] = mx;
for(int e = adj[u]; e; e = nxt[e]){
int v = go[e];
if(v == f) continue;
dfs(now, v, u, max(mx, len[e]));
}
}
}mst;
namespace IO{
inline int read(){
int i = 0, f = 1; char ch = getchar();
for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
if(ch == '-') f = -1, ch = getchar();
for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
return i * f;
}
inline void wr(int x){
if(x < 0) x = -x, putchar('-');
if(x > 9) wr(x / 10);
putchar(x % 10 + '0');
}
}using namespace IO;
int main(){
freopen("h.in", "r", stdin);
n = read(), m = read(), k = read();
ans2 = OO;
mst.init();
for(int i = 1; i <= m; i++)
edge[i].x = read(), edge[i].y = read(), edge[i].t = read();
ans1 = mst.kruskals();
for(int i = 1; i <= n; i++)
mst.dfs(i, i, 0, 0);
// for(int i = 1; i <= n; i++)
// for(int j = 1; j <= n; j++)
// cout<<i<<" "<<j<<" "<<mst.d[i][j]<<endl;
for(int i = 1; i <= m; i++){
int x = edge[i].x, y = edge[i].y, t = edge[i].t;
if(mst.used[i] == 1)
ans2 = min(ans2, ans1 - edge[i].t + (k - edge[i].t));
else
ans2 = min(ans2, ans1 - mst.d[x][y] + (k - edge[i].t));
}
printf("%d", ans2);
return 0;
}