HDU 4126 Genghis Khan the Conqueror （树形DP + MST）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4126

题意：给n个点，m条边，每条边权值c，现在要使这n个点连通。现在已知某条边要发生突变，再给q个三元组，每个三元组(a,b,c)，(a,b)表示图中可能发生突变的边，该边一定是图中的边。c表示该边新的权值，c只可能比原来的权值大。给的q条边发生突变的概率是一样的。求突变后连通n个点最小代价期望值。

思路：上次就看到了这个题，这次又好好做了一下，确实是很好的题，思路来自：http://blog.csdn.net/ophunter_lcm/article/details/12030593

跑出MST后，对其上的边进行dfs,每次都是根据割边枚举含i和不含i的子树之间的最短距离，时间复杂度为O(N^2)

#include <cstdio>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <functional>#include <utility>#include <cmath>#include <cstring>using namespace std;const int maxn = 3010;const int maxm = 3010 * 3010;const int inf = 0x3f3f3f3f;int n, m, q;int d[maxn][maxn];int used[maxn][maxn];int dp[maxn][maxn];vector <int> g[maxn];struct edge{    int from, to, w, nxt;    edge() {}    edge(int from, int to, int w) : from(from), to(to), w(w) {}    bool operator < (const edge & rhs) const    {        return w > rhs.w;    }};struct Prim{    int head[maxn], dis[maxn], vis[maxn];    int dep[maxn], pre[maxn], cnt;    int n, sum;    edge e[maxm];    void init(int n)    {        this -> n = n;        cnt = 0;        sum = 0;        memset(head, -1, sizeof(head));    }    void add(int u, int v, int w)    {        e[cnt].from = u;        e[cnt].to = v;        e[cnt].w = w;        e[cnt].nxt = head[u];        head[u] = cnt++;    }    void prim(int s)    {        priority_queue <edge> que;        memset(dis, inf, sizeof(dis));        memset(vis, 0, sizeof(vis));        memset(dep, 0, sizeof(dep));        memset(pre, -1, sizeof(pre));        vis[s] = 1;        dis[s] = 0;        for(int i = head[s]; ~i; i = e[i].nxt)        {            int v = e[i].to;            dis[v] = e[i].w;            que.push(edge(s, v, e[i].w));        }        while(!que.empty())        {            edge now = que.top();            que.pop();            int u = now.to;            if(vis[u])                continue;            vis[u] = 1;            used[now.from][now.to] = used[now.to][now.from] = 1;        //这条边是不是MST上的边            pre[u] = now.from;            dep[u] = dep[now.from] + 1;            dis[u]  = now.w;            sum += now.w;            for(int i = head[u]; ~i; i = e[i].nxt)            {                int v = e[i].to;                int w = e[i].w;                if(!vis[v] && dis[v] > w)                {                    dis[v] = e[i].w;                    que.push(edge(u, v, w));                }            }        }    }} mst;void init(int n){    memset(used, 0, sizeof(used));    memset(d, inf, sizeof(d));    memset(dp, inf, sizeof(dp));    for(int i = 0; i <= n; i++)    {        d[i][i] = 0;        g[i].clear();    }}int dfs(int cur, int u, int fa)     //从cur点来更新cur所在子树和另一子树的最短距离{    int ans = inf;    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(v == fa) continue;        int tmp = dfs(cur, v, u);       //从cur点来更新以u, v为割边的两子树最短距离        dp[u][v] = dp[v][u] = min(dp[u][v], tmp);        ans = min(tmp, ans);        //以fa, u为割边的两子树最短距离    }    if(fa != cur)                   //生成树边不更新，因为是以该边为割边        ans = min(ans, d[u][cur]);    return ans;}int main(){    while(~scanf("%d%d", &n, &m) && (n + m))    {        mst.init(n);        init(n);        for(int i = 0; i < m; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            mst.add(u, v, w);            mst.add(v, u, w);            d[u][v] = d[v][u] = w;        }        int s = 0;        mst.prim(s);        int ans = 0;        int sum = mst.sum;        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                if(used[i][j])                    g[i].push_back(j);        for(int i = 0; i < n; i++)            dfs(i, i, -1);        scanf("%d", &q);        for(int i = 0; i < q; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            if(used[u][v])            {                if(w < dp[u][v])                    ans += (sum - d[u][v] + w);                else                    ans += (sum - d[u][v] + dp[u][v]);            }            else                ans += sum;        }        printf("%.4f\n", 1.0 * ans / q);    }    return 0;}