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题意：给一图，n个点，m条边，每条边有个花费，给出q条可疑的边，每条边有新的花费，每条可疑的边出现的概率相同，求不能经过原来可疑边
（可以经过可疑边新的花费构建的边），注意每次只出现一条可疑的边，n个点相互连通的最小花费的期望。
析：要想连通先让他们连通起来，先构造出一个MST，然后再暴力，如果这个边不在这里面，那么花费不变，如果在里面，那我们需要知道是用原来的边最少，
还是再找一条边使他们连通起来，这里就要先预处理了，dp[i]j[i] 表示 左边的那个一半 i 和 右边那一半 j 的最长距离，如果我们知道了，就可以用这个来比较了，
我们要对MST进行 n 次更新，每次遍历是 n，所以时间复杂度是 O(n*n)，可以实现。
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 <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e3 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
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 int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
int u, v, val;
Node() { }
Node(int uu, int vv, int va) : u(uu), v(vv), val(va) { }
bool operator < (const Node &p) const {
return val < p.val;
}
};
struct Edge{
int to, next;
};
Edge edge[maxn<<1];
Node a[maxn*maxn];
int p[maxn], dist[maxn][maxn], head[maxn];
int dp[maxn][maxn];
bool is_tree[maxn][maxn];
int cnt, sum;
int Find(int x) { return x == p[x] ? x : p[x] = Find(p[x]); }
void add(int u, int v){
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
}
void Kruskal(){
sort(a, a+m);
int cnt = 0;
sum = 0;
for(int i = 0; i < m; ++i){
int x = Find(a[i].u);
int y = Find(a[i].v);
if(x != y){
p[y] = x;
add(a[i].u, a[i].v);
add(a[i].v, a[i].u);
is_tree[a[i].u][a[i].v] = is_tree[a[i].v][a[i].u] = true;
sum += a[i].val;
++cnt;
}
if(cnt == n-1) break;
}
}
int dfs(int u, int fa, int root){
int ans = fa == root ? INF : dist[root][u]; //i d scf h
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
int tmp = dfs(v, u, root);
ans = Min(ans, tmp);
dp[u][v] = dp[v][u] = Min(dp[u][v], tmp);
}
return ans;
}
int main(){
while(scanf("%d %d", &n, &m) == 2 && m+n){
for(int i = 0; i < n; ++i) p[i] = i;
int u, v, c;
memset(dist, INF, sizeof dist);
for(int i = 0; i < m; ++i){
scanf("%d %d %d", &u, &v, &c);
a[i] = Node(u, v, c);
dist[u][v] = dist[v][u] = c;
}
memset(head, -1, sizeof head);
memset(is_tree, false, sizeof is_tree);
cnt = 0;
Kruskal();
memset(dp, INF, sizeof dp);
for(int i = 0; i < n; ++i) dfs(i, -1, i);
scanf("%d", &m);
double ans = 0.0;
for(int i = 0; i < m; ++i){
scanf("%d %d %d", &u, &v, &c);
if(!is_tree[u][v]) ans += sum;
else ans += sum - dist[u][v] + Min(c, dp[u][v]);
}
printf("%.4f\n", ans/m);
}
return 0;
}
HDU 4126 Genghis Khan the Conqueror (树形DP+MST)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5935004.html
