Building Forest CodeForces, forestcodeforces

Building Forest CodeForces, forestcodeforces

Building Forest CodeForces-195E

This is really hard to understand... It means "An oriented weighted forest is an acyclic weighted digraph in which from each vertex at most one edge goes." What does this sentence mean...

N vertices need to be inserted into a forest with edge weights in order of 1-n (the forest does not have any points or edges at the beginning ), line I + 1 describes the operation after point I is inserted. If a row is (K, Zookeeper,V1. zookeeper,X1. zookeeper,V2, Zookeeper,X2, please wait..., please wait ,...,VK, Zookeeper,XK), Then k indicates the number of edges to be connected. (vj, xj) indicates that an edge is to be connected to I from vertex vj, its weight is the total length of the path from vj to the root node of the original tree plus xj. (Ensure that data does not produce duplicate edges)

Practice: After understanding the question, you will find that this is a very normal (shui) weighted query set.

1 # include <cstdio> 2 # define md 1000000007 3 typedef long LL; 4 LL fa [100100]; 5 LL hei [100100]; // The total path length from record point I to Father's Day is 6 LL ans, n; 7 LL find (LL x) 8 {9 if (fa [x] = x) return x; 10 LL t = find (fa [x]); 11 hei [x] = (hei [fa [x] + hei [x]) % md; 12 fa [x] = t; 13 return fa [x]; 14} 15 int main () 16 {17 LL I, j, k, v, x, f1; 18 scanf ("% I64d", & n); 19 for (I = 1; I <= n; I ++) 20 fa [I] = I; 21 for (I = 1; I <= n; I ++) 22 {23 scanf ("% I64d", & k); 24 for (j = 1; j <= k; j ++) 25 {26 scanf ("% I64d % I64d", & v, & x); 27 f1 = find (v ); 28 // if (f1 = I) continue; 29 fa [f1] = I; 30 hei [f1] = (x + hei [v]) % md; 31 ans = (ans + hei [f1]) % md; 32} 33} 34 printf ("% I64d", (ans + md) % md); 35 return 0; 36}

Mistakes made:

1. The answer is not modulo, and a negative number is output (twice written, twice wrong)

2. Misunderstanding: 12 lines

1 # include <cstdio> 2 # define md 1000000007 3 typedef long LL; 4 LL fa [100100]; 5 LL hei2 [100100]; // The weight of the edge of the record point I connected to his father's day point is 6 LL hei [100100]; // The total length of the path from the record point I to the Father's Day point is 7 LL ans, n; 8 LL find (LL x) 9 {10 if (fa [x] = x) return x; 11 LL t = find (fa [x]); 12 hei [x] = (hei [fa [x] + hei2 [x]) % md; 13 fa [x] = t; 14 return fa [x]; 15} 16 int main () 17 {18 LL I, j, k, v, x, f1; 19 scanf ("% I64d", & n ); 20 for (I = 1; I <= n; I ++) 21 fa [I] = I; 22 for (I = 1; I <= n; I ++) 23 {24 scanf ("% I64d", & k); 25 for (j = 1; j <= k; j ++) 26 {27 scanf ("% I64d % I64d", & v, & x); 28 f1 = find (v); 29 if (f1 = I) continue; 30 fa [f1] = I; 31 hei2 [f1] = (x + hei [v]) % md; 32} 33} 34 for (I = 1; I <= n; I ++) 35 ans = (ans + hei2 [I] + md) % md; 36 printf ("% I64d", ans); 37 return 0; 38}