Topic Links:
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1804
Main topic:
A directed acyclic graph (DAG) that has N points M bars have a forward edge (n,m<=105), each point has two values Ai,bi (ai,bi<=109), and Count (I,j) represents the number of scenarios that go from I to J.
Find the value of mod 109+7.
Topic Ideas:
"Topology" "Wide Search"
First, the equation is opened, each point I go to the J D[j] Once added AI, such a point by I to several times to add a few AI, the equivalent of Count (i,j) *ai, ultimately only required σ (Bj*d[j]) can.
Add this forward-free graph to the queue according to the topological order, so as to ensure that the second point is not updated after the answer (that is, it will not affect the previous value, so that the previous value can be pushed back)
Each point I will be the point I of the successor node D plus AI, and finally in accordance with the above to seek σ (Bj*d[j]) can be.
1 //2 //by Coolxxx3 //#include <bits/stdc++.h>4#include <iostream>5#include <algorithm>6#include <string>7#include <iomanip>8#include <map>9#include <stack>Ten#include <queue> One#include <Set> A#include <bitset> -#include <memory.h> -#include <time.h> the#include <stdio.h> -#include <stdlib.h> -#include <string.h> - //#include <stdbool.h> +#include <math.h> - #defineMin (a) < (b) ( A):(B)) + #defineMax (a) (a) > (b)? ( A):(B)) A #defineABS (a) ((a) >0? ( A):(-(a))) at #defineLowbit (a) (a& (a)) - #defineSqr (a) ((a) * (a)) - #defineSwap (a) (a) ^= (b), (b) ^= (a), (a) ^= (b)) - #defineMem (A, B) memset (A,b,sizeof (a)) - #defineEPS (1E-8) - #defineJ 10000 in #defineMoD 1000000007 - #defineMAX 0x7f7f7f7f to #definePI 3.14159265358979323 + #defineN 100004 - using namespacestd; thetypedefLong LongLL; * intCas,cass; $ intN,m,lll,ans;Panax Notoginseng LL Aans; - structXXX the { + intnext,to; A }a[n]; the intLast[n],q[n],inch[n],aa[n],bb[n]; + LL D[n]; - voidAddintXinty) $ { $a[++lll].to=y; -a[lll].next=Last[x]; -last[x]=lll; the } - voidTuopu ()Wuyi { the intI,l,r=0; - for(i=1; i<=n;i++)if(!inch[i]) q[++r]=i; Wu for(l=1; L<=r && r!=n;l++) - for(i=last[q[l]];i;i=a[i].next) About if(--inch[a[i].to]==0) $q[++r]=a[i].to; - for(l=1; l<=n;l++) - for(i=last[q[l]];i;i=a[i].next) -D[a[i].to]= (D[a[i].to]+d[q[l]]+aa[q[l])%MoD; A for(i=1; i<=n;i++) +Aans= (Aans+1ll*d[i]*bb[i])%MoD; the } - intMain () $ { the #ifndef Online_judge theFreopen ("1.txt","R", stdin); the //freopen ("2.txt", "w", stdout); the #endif - inti,j,k; in intx, y, z the //for (scanf ("%d", &cass); cass;cass--) the //for (scanf ("%d", &cas), cass=1;cass<=cas;cass++) About //while (~scanf ("%s", s+1)) the while(~SCANF ("%d",&N)) the { thelll=aans=0; Mem (Last,0); Mem (inch,0); Mem (d,0); +scanf"%d",&m); - for(i=1; i<=n;i++) scanf ("%d%d", aa+i,bb+i); the for(i=1; i<=m;i++)Bayi { thescanf"%d%d",&x,&y); theAdd (x, y);inch[y]++; - } - Tuopu (); theprintf"%lld\n", Aans); the } the return 0; the } - /* the // the the //94 */
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Topology "wide search" CSU 1084 with Direction-free graph (2016 Hunan province 12th session of university Students computer Program design contest)