'commit'

TangDou/AcWing_TiGao/T5/QiWang/217.cpp

@@ -16,7 +16,7 @@ double dfs(int u) {
     f[u] = 0;                   // 初始化为0，准备开始填充
     for (int i = h[u]; ~i; i = ne[i]) {
         int j = e[i];
-        f[u] += (w[i] + dfs(j)) / out[u]; // 看推的公式
+        f[u] += (w[i] + dfs(j)) / out[u];
     }
     return f[u];
 }

TangDou/AcWing_TiGao/T5/QiWang/217.drawio

@@ -1 +0,0 @@
-<mxfile host="Electron" modified="2022-06-20T07:53:04.288Z" agent="5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) draw.io/16.0.2 Chrome/96.0.4664.55 Electron/16.0.5 Safari/537.36" etag="jNveJB-z9TWn2FKd464S" version="16.0.2" type="device"><diagram id="S7yhPZ9SSO8EGXcPaCVl" name="第 1 页">5VpLc9owEP41zLSHMtbLj2NDHj00M+mkbdLePFiAEmNRI4rJr6+MZbAlB0gCVgkzHLwraWV9+3m1WtFBvXF2lYaT0TWPaNyBTpR10HkHQoAh7OQ/J1oUGt8jhWKYskh1Witu2RNVSkdpZyyi01pHwXks2KSu7PMkoX1R04Vpyuf1bgMe12edhENqKG77YWxq71gkRkrrErxu+ELZcFRODdygaBmHZW+1lOkojPi8okIXHdRLORfF0zjr0ThHrwSmGHf5TOvqzVKaiF0GDB7cy3F2+fPp4QpP/izI7Ovi/pOy8jeMZ2rFD+ptxaLEQJqRcEvhbD5igt5Own7eMpcel7qRGMdSAvIxnE4KHwxYRuWsZ1OR8scVcFhqBjwRyssQlbKaqmFF5evRVNCsolIrvKJ8TEW6kF1Ua4AV2opvIFDyfO09olSjit9KXaj4MlxZXiMqHxSozQBf9JxfT78ZFL9/XH0PxfUw/PatAeAM2EbYgLMB9GcRdoM6wtCxjDA0EUZHjbDneTWEEbKMMDIRhkeNcKBxGNvmMDYRfjxqhH2/jjCxzWFiIJwcNcAu1AC2TWHXANjEN4k+5zmZlPpxOJ2yfh3WlM+SKMfz3NEQ9KRMMybuVVv+/Ct/7hIlnWeVpvNFKSRyaffK/FKojMrF9bClVI4z/VkshkZGumgkK1M+S/t0e84lwnRIxbbUwSTAjg5OaRwK9rf+uk1eVzPccCYXst6EHC1EehpximWqUdW0UzdENENQM1TgYBhaknC17Nfz0jN4edeQgckPUdTJWFCgx2OeSk3CkzwuDFgca6owZsMkZ7TkAZX6s/yzZvIE8Vk1jFkUxc8Flc2U30cg1uIE8M044TbQCB0qTvitxYku9CuhArwmULw4TORGb2jKJFY5GZa9LIQOaDN0BNoXLw3vJ3RApBk6cOgI2qMq2Zmp/8XWhK3yy8V1Wug5za788n3YdT3PdQMijWpWEfa6aHn4klsfQq7jtsq9sgB2ePKBN1BvFTYrKdUugdMCZZFVypZlz7dSNgBBTlnfgW7g5QyumSXY6WIMCCI+CggO2mWsWeo6VLh8wQlgA2e1MwHcwtmltJ+9fVOVZSuRiVUiBxtCpsztq40owK8jOXBg/XPBoN0zAzBrinum8u603HZU3SMD8TEwEDj6yVTPCnelmavXqFo+mQKzrnqogPmC/PK1W/q+8tJNtyj/NzNPIy81S9Ut7PJvyEs9WI+nEi+7+zw8Bi6/oyhrVv7vGi4IT6YACD3LBUBg3hTcNdwnvluHGJ8Wari6adcjTTVy8/7x3XrE9zSHNPxppF2HmEXywYcMfDwdl7hAcwm27RKzGCxdgk7IJZ7216rVRZ4tl5Tz110CT8gl+t6Obe/t0KwBSpc8npBL9DsrQg7mEimu/z1apMvrP+Gii38=</diagram></mxfile>

TangDou/AcWing_TiGao/T5/QiWang/217.md

@@ -72,22 +72,26 @@ $f(i)$：从$i$跳到$N$的期望长度。边界$f(N)=0$
 
 ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/202312191619342.png)
 
-事件发生的期望的线性性 $$\large E(aX+bY)=aE(X)+bE(Y)$$
 
+题意:给定一个$DAG$，求路径和。
 
-$f[i]$: 从 $i$ 跳到 $N$ 的期望长度
-边界: $f[N]=0$
-答案: $f[1]$
+由题意知，题目中的$K$就是每个点的出度。
 
-![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/202312191621193.png)
+然后，又由题意知，最后一个点，它没有出度，而到自己的距离为$0$，即状态已经确定。
 
-则有如下递推式：
+用前向星存图，从第一个点开始
 
-$\large f(i)=E(\frac{1}{k}(w_1+x_1)+\frac{1}{k}(w_2+x_2)+⋯+\frac{1}{k}(w_k+x_k)) \\
-=E(\frac{1}{k}(w_1+x_1))+E(\frac{1}{k}(w_2+x_2))+⋯+E(\frac{1}{k}(w_k+x_k))\\ 
-=\frac{1}{k}((w_1+E(x_1))+(w_2+E(x_2))+⋯+(w_k+E(x_k))) \\
-=\frac{1}{k}((w_1+f(i_1))+(w_2+f(i_2))+⋯+(w_k+f(i_k)))$
+$DFS$,每次枚举它所连的边，并把边的边权 ($w[i]$)加上，最后除以它的出度即可。
 
+方程： 
+$dp[n]=0 \\
+dp[i]+=(dp[son[i]]+e[son[i]].val)/out[i]$
+
+其中，
+
+$son[i]$是$i$所连得边，$e$是前向星，$val$是边权,$out[i]$是$i$的出度。
+
+#### $Code$
 ```cpp {.line-numbers}
 #include <bits/stdc++.h>
 using namespace std;
@@ -107,7 +111,7 @@ double dfs(int u) {
     f[u] = 0;                   // 初始化为0，准备开始填充
     for (int i = h[u]; ~i; i = ne[i]) {
         int j = e[i];
-        f[u] += (w[i] + dfs(j)) / out[u]; // 看推的公式
+        f[u] += (w[i] + dfs(j)) / out[u]; 
     }
     return f[u];
 }

TangDou/AcWing_TiGao/T5/QiWang/217_DaoTui.cpp

@@ -1,43 +0,0 @@
-#include <bits/stdc++.h>
-using namespace std;
-const int N = 100010, M = N << 1;
-int n, m;
-int in[N], g[N]; // 入度，入度的备份数组，原因：in在topsort中会不断变小受破坏
-double f[N];
-
-// 链式前向星
-int e[M], h[N], idx, w[M], ne[M];
-void add(int a, int b, int c = 0) {
-    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
-}
-
-void topsort() {
-    queue<int> q;
-    q.push(n);
-    f[n] = 0; // n到n的距离期望是0
-    while (q.size()) {
-        int u = q.front();
-        q.pop();
-        for (int i = h[u]; ~i; i = ne[i]) { // 枚举每条入边(因为是反向图)
-            int v = e[i];
-            f[v] += (f[u] + w[i]) / g[v];
-            in[v]--;
-            if (in[v] == 0) q.push(v);
-        }
-    }
-}
-int main() {
-    memset(h, -1, sizeof h);
-    cin >> n >> m;
-
-    while (m--) {
-        int a, b, c;
-        cin >> a >> b >> c;
-        add(b, a, c); // 反向图,计算从n到1
-        in[a]++;      // 入度
-        g[a] = in[a]; // 入度数量
-    }
-    topsort();
-    printf("%.2lf\n", f[1]);
-    return 0;
-}

TangDou/AcWing_TiGao/T5/QiWang/217_ZhengTui.cpp

@@ -1,59 +0,0 @@
-#include <bits/stdc++.h>
-using namespace std;
-const int N = 1e5 + 10, M = N << 1;
-
-// 邻接表
-int h[N], e[M], ne[M], w[M], idx;
-void add(int a, int b, int c) {
-    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
-}
-
-int n, m; // n个顶点，m条边
-
-int out[N], in[N]; // 出度,入度
-double f[N], g[N]; // f:数学期望结果 g:概率
-
-void topsort() {
-    queue<int> q;
-    // 起点为1,起点的概率为100%
-    q.push(1);
-    g[1] = 1.0;
-    f[1] = 0.0;
-
-    // DAG，执行拓扑序,以保证计算的顺序正确，确保递归过程中，前序数据都已处理完毕
-    while (q.size()) {
-        auto u = q.front();
-        q.pop();
-
-        for (int i = h[u]; ~i; i = ne[i]) { // 枚举的是每边相邻边
-            int v = e[i];                   // 此边，一端是t，另一端是j
-            // 此边边条w[i]
-            f[v] += (f[u] + w[i] * g[u]) / out[u];
-            g[v] += g[u] / out[u]; // g[j]也需要概率累加
-            // 拓扑序的标准套路
-            in[v]--;
-            if (!in[v]) q.push(v);
-        }
-    }
-}
-
-int main() {
-    // 初始化邻接表
-    memset(h, -1, sizeof h);
-    cin >> n >> m;
-
-    while (m--) {
-        int a, b, c;
-        cin >> a >> b >> c;
-        add(a, b, c);
-        // 维护出度，入度
-        out[a]++, in[b]++;
-    }
-    // 拓扑序
-    topsort();
-
-    // 正向递推，输出结果，保留两位小数
-    printf("%.2lf", f[n]);
-
-    return 0;
-}