diff --git a/8.15/F-Candies/F-Candies.cpp b/8.15/F-Candies/F-Candies.cpp
new file mode 100644
index 0000000..b27b54f
--- /dev/null
+++ b/8.15/F-Candies/F-Candies.cpp
@@ -0,0 +1,70 @@
+#include <iostream>
+#include <cstdio>
+#include <algorithm>
+#include <cstring>
+#include <stack>
+using namespace std;
+const int N = 3e4 + 5, N3 = 15e4 + 5, inf = 0x3f3f3f3f;
+int to[N3], ne[N3], w[N3], h[N], cnt = 0, s = N - 4, e, dist[N];
+bool in[N], wa = 0;
+int  u, v, w1;//WA 和bool放一起了
+int n, m, maxn = 0;
+void add(int u, int v, int w1)
+{
+	to[++cnt] = v;
+	ne[cnt] = h[u];
+	w[cnt] = w1;
+	h[u] = cnt;
+}
+void spfa(int s)
+{
+	memset(dist,0x3f,sizeof dist);//更好的賦初值方法
+	dist[s] = 0;//和賦inf顺序反了=白干
+	stack<int> q; // queue没有自带函数清零
+	q.push(s);
+	while (!q.empty())
+	{
+		u = q.top();
+		q.pop();
+		in[u] = 0; // spfa的精髓！！
+		for (int i = h[u]; i; i = ne[i])
+		{
+			v = to[i], w1 = w[i];
+			//			printf("%d %d %d\n",u,v,w1);
+			if (dist[u] + w1 < dist[v])
+			{
+				dist[v] = dist[u] + w1;
+				//			printf("%d(%d) %d(%d) %d\n",u,dist[u],v,dist[v],w1);
+				if (!in[v])
+				{
+					q.push(v);
+					in[v] = 1;
+				}
+			}
+		}
+	}
+}
+int main()
+{
+	scanf("%d%d", &n, &m);
+	// for (int i = 0; i <= n; i++)
+	// {
+	// 	dist[i] = inf;
+	// }
+	for (int i = 1; i <= m; i++) // n,m不分
+	{
+		scanf("%d%d%d", &u, &v, &w1);
+		add(u, v, w1);
+	}
+	spfa(1);
+	// maxn = 0;
+	//        for(int i=2;i<=n;i++)
+	//        {
+	//        	if(dist[i]>maxn)
+	//        	{
+	//        		maxn=dist[i];
+	//			}
+	//		}
+	printf("%d", dist[n]);
+	return 0;
+}
diff --git a/8.15/F-Candies/Readme.md b/8.15/F-Candies/Readme.md
new file mode 100644
index 0000000..2c550fb
--- /dev/null
+++ b/8.15/F-Candies/Readme.md
@@ -0,0 +1,17 @@
+# F - Candies
+* https://vjudge.net/contest/509525#problem/F
+### 题意
+给定m对A认为B可以比他大C,求编号1与编号n的最大差距
+### 做法
+差分约束、单源最短路
+### 关键词
+差分约束、sfpa、inf、memset
+### 易错点
+//WA！ int和bool放一起了
+dist[s] = 0;//和賦inf顺序反了=白干
+stack<int> q; // queue没有自带函数清零
+in[u] = 0; // spfa的精髓！！不要忘了每次弹出
+// n,m不分
+### 工具箱
+* memset(dist,0x3f,sizeof dist);//更好的賦初值方法 https://www.cnblogs.com/td15980891505/p/5431898.html
+* queue<int> q; // queue没有自带函数清零,每次重开就行
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