The problem of two topological sorts

How long has not written the thing ah ....

Two topological sorting Liv.1 .... The method is the same ~ ~

"Topology sequencing • Two"

Title: HTTP://HIHOCODER.COM/CONTEST/HIHO81/PROBLEM/1

A computer network, one-way edge, if there is an edge u->v, then the virus of u will infect v.

Point, there is no ring! Then if you =0, then the number of viruses in u will not change.

Think of the topology sort. Continue to delete the point of 0 in degrees. Each time you delete node U, if there is u->v, then the number of viruses

NUM[V] + = Num[u]. Problem solving.

(using queue to achieve topological ordering is very convenient ~ ~)

Code:

1 #include <bits/stdc++.h>23UsingNamespaceStd4Constint MAXN =100000 +10;5Constint mod =142857;6IntN, M, K;7int NUM[MAXN],Inch[MAXN];8 vector<Int>G[MAXN];910IntSolve () {queue<Int>Q1213int ans =0;14Forint i =1; I <= N; ++i) {15if (!Inch[i]) {16//cout << "I:" << i << Endl;17Q.push (i);18}19}2021stwhile (!Q.empty ()) {22int u =Q.front (); Q.pop ();% ans = (ans + num[u])%MoD2425Forint i =0; I < g[u].size (); ++i) {26int v =G[u][i];27in[v]--; NUM[V] = (Num[u] + num[v])%MoD28if (!Inch[v])29Q.push (v);30}31}3233ReturnAns34}3536IntMain () {Panax Notoginseng scanf ("%d%d%d", &n, &m, &k);38The memset (NUM,0,sizeof(num));40IntU, v;41Forint i =0; I < K; ++i) {scanf ("%d", &u);num[u]++;44}45memset (In0,sizeofInch) );47Forint i =0; I < m; ++i) {scanf ("%d%dv) ; 49 G[u].push_back (v); in[v]++; }51 52 int ans = solve (); %d\n" , ans); 54 55 return 0; *                 

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"Hiho for the 81st Week"

Title: HTTP://HIHOCODER.COM/CONTEST/HIHO81/PROBLEM/1

method is the same, the map is slightly difficult ...

Code:

1 #include <bits/stdc++.h>23UsingNamespaceStd4Constint MAXN =100000 +10;5Constint mod =142857;6IntN, M;7 vector<Int>Inch[MAXN];8IntS[MAXN], RU[MAXN], NUM[MAXN];9 vector<Int>E[MAXN], G[MAXN];1011voidInit () {12Forint i =0; i < MAXN; ++I13Inch[I].clear (), E[i].clear (), g[i].clear ();memset (s),0,sizeof(s));memset (Ru,0,sizeof(RU));memset (NUM,0,sizeof(num));17}1819voidSolve () {queue<Int>Q21stForint i =0; I <= N; ++i) {22if (!Ru[i])23Q.push (i);24}2526while (!Q.empty ()) {27int u =Q.front (); Q.pop ();28Forint i =0; I < g[u].size (); ++i) {29int v =G[u][i];RU[V]--, Num[v] = (Num[v] + num[u])%MoD31if (!RU[V])32Q.push (v);33}34}35}3637voidPrint () {* printf ("%d", num[1]);39Forint i =2; I <= N; ++i) {printf ("%d", Num[i]);41}printf ("\ n");43}4445IntMain () {46//Freopen ("1.in", "R", stdin);47IntTscanf ("%d", &T);4950while (t--){51Init ();scanf ("%d%d", &n, &m);5354IntU, V, K;s[0] =-1, num[0] =1;56Forint i =0; I < m; ++i) {scanf ("%d", &u);e[0].push_back (U);59}6061Forint i =1; I <= N; ++i) {scanf ("%d%d", &u, &k);S[i] = u;Inch[U].push_back (i);64ForInt J =0; J < K; ++j) {scanf ("%d", &u);66E[i].push_back (U);67}68}6970//Build graph71Forint i =0; I <= N; ++i) {72ForInt J =0; J < E[i].size (); ++j) {73int sg =E[I][J];74Forint p =0; P <In[sg].size (); ++P) {75int v =IN[SG][P]; ru[v]++;76  G[i].push_back (v) }78 }79 }80 //output (); 81 82  solve (); 83  print (); }85 86 return 0;                

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The problem of two topological sorts