题意:给定一棵树,让你选一个中心城市,问你中心城市到所有其他城市的权值和最小,每条路的权值是该边上和最小权值。
析:从大到小枚举边,然后用并查集进行维护,在合并两个集合时,要考虑两边集合加上该边的的最大的那一个,每次要取最大值。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2e5 + 10; const LL mod = 1e12; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int p[maxn]; int num[maxn]; LL sum[maxn]; struct Node{ int u, v, w; bool operator < (const Node &p) const{ return w > p.w; } }; Node a[maxn]; int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); } int main(){ while(scanf("%d", &n) == 1){ for(int i = 0; i < n-1; ++i) scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w); memset(sum, 0, sizeof sum); for(int i = 0; i <= n; ++i){ p[i] = i; num[i] = 1; } sort(a, a + n - 1); for(int i = 0; i < n-1; ++i){ int x = Find(a[i].u); int y = Find(a[i].v); LL tmp1 = sum[x] + (LL)num[y] * a[i].w; LL tmp2 = sum[y] + (LL)num[x] * a[i].w; sum[x] = max(tmp1, tmp2); num[x] += num[y]; p[y] = x; } printf("%I64d ", sum[Find(1)]); } return 0; }