CSU 1804: A topological sort diagram of a direction-free graph

1804: A direction-free graph

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The Descriptionbobo has an n point, a forward-free graph of M-bars (i.e., for any point V, there is no path from point V to end of Point v). For convenience, click,..., N number. Set count (x, y) to indicate the number of paths (specified by count (x,x) =0) for points x to Y, Bobo want to know the remainder of the divide (109+7). Where AI,BJ is a given number of sequences. Input inputs contain no more than 15 sets of data. The first row of each group of data contains two integers n,m (1≤n,m≤105). The next line of n lines contains two integers, Ai,bi (0≤ai,bi≤109). The last M line contains two integers ui,vi, representing an edge from the point UI to the VI (1≤UI,VI ≤n). Output for each set of data, outputs an integer that represents the required value. Sample Input

3 31 11 11 11 21 32 32 21 00 21 21 22 1500000000 00 5000000001 2

Sample Output

44250000014

Hintsource Hunan Province The 12th session of the computer Program design competition to calculate each point I corresponding to the count (i,j) *BJ and then by AI, accumulation is the answer, note here to be similar to the topological sort, but to do backwards, to avoid the aftereffect of Reference blog HTTP/ blog.csdn.net/qq_21057881/article/details/52431139

#include <stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<vector>#include<queue>using namespacestd;#defineLL Long LongConst intMAXN = 1e5+7;Const intMoD = 1e9+7; Vector<int>E[MAXN];intA[MAXN],B[MAXN],D[MAXN];intANS[MAXN];intMain () {intn,m;  while(Cin >> N >>m) {         for(intI=1; i<=n;i++) {e[i].clear (); } memset (d,0,sizeof(d)); memset (ans,0,sizeof(ans));  for(intI=1; i<=n;i++) {cin>> A[i] >>B[i]; }         for(intI=1; i<=m;i++){            intu,v; CIN>> u >>v;            E[v].push_back (U); D[u]++;//number of paths to the end of U} queue<int>Q;  for(intI=1; i<=n;i++){            if(D[i] = =0){//Join the queue with a point with a path number of 0 at the end of IQ.push (i); }        }         while(!Q.empty ()) {            intv =Q.front ();            Q.pop ();  for(intI=0; I<e[v].size (); i++){                intU =E[v][i]; Ans[u]= (Ans[u] + (Ans[v] + b[v])%mod)%MoD; //The reason is add b[v], is because the multiplication is equivalent to the whole, one is 1*b[v] equivalent to add b[v]d[u]--; if(D[u] = =0) {Q.push (U); }}} LL res=0;  for(intI=1; i<=n;i++) {res= (res + 1ll*ans[i]*a[i]%mod)%MoD; } cout<< Res <<Endl; }    return 0;}

CSU 1804: A topological sort diagram of a direction-free graph