Wed, 30 Jan 2019 00:05:38 +0800

C279991/.tutorial.md

@@ -23,3 +23,7 @@
 ### F - Kiki & Little Kiki 1
 
 multiset.upper_bound()
+
+### G - 传递
+
+题意看起来非常奇怪，但是实际上只要用bitset存一下一个点连向了哪些点即可。只要保证对于每一条边，其终点所连向的点的集合为起点连向的点的集合的子集即可判断这个图是“传递”的。

C279991/CMakeLists.txt

@@ -8,4 +8,5 @@ add_executable("C" "C.cpp")
 add_executable("D" "D.cpp") 
 add_executable("E" "E.cpp") 
 add_executable("F" "F.cpp") 
+add_executable("G" "G.cpp") 
 add_executable("Z" "Z.cpp") 

C279991/G.cpp

@@ -0,0 +1,33 @@
+#define _SILENCE_CXX17_C_HEADER_DEPRECATION_WARNING
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1 << 11;
+bitset<N> a[2][N];
+char buf[N];
+int main()
+{
+    int T, n;
+    scanf("%d", &T);
+    while (T--)
+    {
+        memset(a, 0, sizeof(a));
+        scanf("%d", &n);
+        for (int i = 0; i < n; i++)
+        {
+            scanf("%s", buf);
+            for (int j = 0; j < n; j++)
+            {
+                if (buf[j] == 'P') a[0][i].set(j);
+                if (buf[j] == 'Q') a[1][i].set(j);
+            }
+        }
+        bool flag = true;
+        for (int k = 0; k < 2 && flag; k++)
+            for (int i = 0; i < n && flag; i++)
+                for (int j = 0; j < n && flag; j++)
+                    if (a[k][i][j] && (a[k][i] | a[k][j]) != a[k][i])
+                        flag = false;
+        puts(flag ? "T" : "N");
+    }
+    return 0;
+}