[2019南京网络赛D题]Robots

题目链接

2019.9.2更新

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第二天睡醒想了想发现好像搜一遍就可以过，赛时写的花里胡哨的还错了，太菜了QAQ

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 3e5 + ;
 struct node {
     int s, e, next;
 }edge[maxn];
 int head[maxn], len;
 void init() {
     memset(head, -, sizeof(head));
     len = ;
 }
 void add(int s, int e) {
     edge[len].s = s;
     edge[len].e = e;
     edge[len].next = head[s];
     head[s] = len++;
 }
 double dp[maxn], dp2[maxn];
 int vis[maxn],n;
 void dfs(int x) {
     if (vis[x])
         return;
     vis[x] = ;
     dp[x] = dp2[x] = ;
     if (x == n)
         return;
     int num = ;
     for (int i = head[x]; i != -; i = edge[i].next) {
         int y = edge[i].e;
         dfs(y);
         dp[x] += dp[y];
         dp2[x] += dp2[y];
         num++;
     }
     dp[x] += num + , dp[x] /= num;
     dp2[x] += (num + )*dp[x], dp2[x] /= num;
 }
 int main() {
     int t;
     scanf("%d", &t);
     while (t--) {
         int  m, x, y;
         init();
         scanf("%d%d", &n, &m);
         for (int i = ; i <= m; i++)
             scanf("%d%d", &x, &y), add(x, y);
         memset(vis, , sizeof(vis));
         dfs();
         printf("%.2f\n", dp2[]);
     }
 }

原文

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绝望ing!!!

搞了3个小时的D,到最后也没过，吃饭的时候突然想到错了，改一改就过了Orz.

遗憾!!错在求了拓扑序,要用a[n]->a[1],结果用成了n->1。要被队友骂死了QAQ

而且样例刚好1-n是拓扑序(

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题意是说，在DAG中从1到n，每次有x/(x+1)的概率到相连的点，1/(x+1)的概率原地不动，(x为相连的点的个数)。每次移动需要一天，每次移动会消耗当前天数数值那么多的能量，求到达n点的期望能量消耗。

一开始就蛮有感觉的，DAG上跑期望dp，先跑期望天数，再跑期望能量，设dp[i]为i点到n点的期望天数，则初始dp[n]=0,转移为$dp[i]=\tfrac{1}{x+1}*\sum dp[j]+\tfrac{1}{x+1}*dp[i]+1$ i->j有边。

然后再跑期望能量，dp2[i]为i到n点的期望能量，初始为dp2[n]=0,转移为$dp2[i]=\tfrac{1}{x+1}*\sum dp2[j]+\tfrac{1}{x+1}*dp2[i]+dp[i]$ i->j有边。

代码极度丑陋，望见谅。

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = 4e5 + ;
 struct node {
     int s, e, next;
 }edge[maxn], edge2[maxn];
 int head[maxn], head2[maxn], len, len2;
 void init() {
     memset(head, -, sizeof(head));
     memset(head2, -, sizeof(head2));
     len = len2 = ;
 }
 void add(int s, int e) {
     edge[len].s = s;
     edge[len].e = e;
     edge[len].next = head[s];
     head[s] = len++;
 }
 void add2(int s, int e) {
     edge2[len2].s = s;
     edge2[len2].e = e;
     edge2[len2].next = head2[s];
     head2[s] = len2++;
 }
 double dp[maxn], dp2[maxn];
 int in[maxn], a[maxn], num[maxn];
 void solve(int n) {
     int lens = ;
     queue<int>q;
     for (int i = ; i <= n; i++)
         if (in[i] == ) q.push(i);
     while (!q.empty()) {
         int p = q.front(); q.pop();
         a[++lens] = p;
         for (int i = head[p]; i != -; i = edge[i].next) {
             int y = edge[i].e;
             in[y]--;
             if (in[y] == )
                 q.push(y);
         }
     }
 }
 int main() {
     int t;
     scanf("%d", &t);
     while (t--) {
         int n, m, x, y;
         scanf("%d%d", &n, &m);
         init();
         for (int i = ; i <= n; i++)
             in[i] = , num[i] = , dp2[i] = dp[i] = ;
         for (int i = ; i <= m; i++)
             scanf("%d%d", &x, &y), add(x, y), add2(y, x), in[y]++;//正向反向各存一图，正向跑拓扑，反向求期望
         solve(n);//求拓扑序
         dp[a[n]] = ;
         for (int x = n; x >= ; x--) {//期望dp日常倒推
             if (a[x] != n) dp[a[x]] += num[a[x]] + , dp[a[x]] /= num[a[x]];//num[i]为与i点相连的点数
             num[a[x]] = ;
             for (int i = head2[a[x]]; i != -; i = edge2[i].next) {
                 int y = edge2[i].e;
                 dp[y] += dp[a[x]];
                 num[y]++;
             }
         }
         dp2[n] = ;
         for (int x = n; x >= ; x--) {
             if (a[x] != n) dp2[a[x]] += (num[a[x]] + ) * dp[a[x]], dp2[a[x]] /= num[a[x]];
             for (int i = head2[a[x]]; i != -; i = edge2[i].next) {
                 int y = edge2[i].e;
                 dp2[y] += dp2[a[x]];
                 num[y]++;
             }
         }
         printf("%.2lf\n", dp2[]);
     }
 }