Nyoj 1208--Water Problem series —————— "DP"

Water Problem Series time limit:MS | Memory limit:65535 KB Difficulty:2

Describe

give you a map, each side has a certain weight, and now let you start at any point, the weight of the side of each walk must be the weight of the last time a large number of cases, ask you can go a few sides?

Input

first an n and M, respectively, representing the number of points and the number of edges
The next M-line, three values per line X,y,val, indicates that X-to-Y has a path with a weighted value of Val. (1<n,m,val<10^5,0<x,y<=n)

Output

the maximum number of edges to output

Sample input

3 31 2 12 3 13 1 16 71 2 13 2 52 4 22 5 22 6 95 4 34 3 4

Sample output

 
 
 
 
 Problem Solving: The topic just seems to be a graph theory, but the subject is the DP thought of the investigation. Defines two arrays Dp[i],g[i] indicates the maximum number of forward edges at which the edge is numbered I, and when the point is I. The transfer equation is: 
 If the edge rights are different: Dp[i]=g[e[i].st]+1,g[e[i].en]=max (G[e[i].en],dp[i]). 
 If the edge rights are the same: {i,i+1,i+2,i+3 ...} Edges in the collection are numbered with the same edge. 
 Dp[i]=g[e[i].st]+1,dp[i+1]=g[e[i+1].st]+1,etc ... 
 G[e[i].en]=dp[i],g[e[i+1].en]=dp[i+1],etc ... 
 for the same side of the same situation, the first set of examples can see the problem, here the operation of the seniors said is the delay, specifically should be the point of delay, on the edge ahead. 
 
 
 

#include <bits/stdc++.h>using namespace Std;const int maxe=1e5+10;struct edge{int st,en; int Val;} E[maxe];int dp[maxe],g[maxe];int inf=1e9;bool cmp (Edge A,edge b) {return a.val<b.val;}    void solve (int n,int m) {memset (dp,0,sizeof (DP));    memset (G,0,sizeof (g));            for (int i=0;i<m;i++) {if (e[i].val<e[i+1].val) {dp[i]=g[e[i].st]+1;        G[e[i].en]=max (G[e[i].en],dp[i]);            }else{int j,t=i;            for (j=i;j<m;j++) {if (e[j].val!=e[j+1].val) break;            } i=j;            for (int k=t;k<=j;k++) {dp[k]=g[e[k].st]+1;            } for (int k=t;k<=j;k++) {G[e[k].en]=max (g[e[k].en],dp[k]); }}}} int max_e=0;//for (int i=1;i<=n;i++) {//Traversal point and edge should be the same effect, but the test data in the background is ... if (g[i]>max_e) {//max_e=g[i];//}//} for (int i=0;i<m;i++) {if (max_ E<dp[i]) {                Max_e=dp[i]; }} printf ("%d\n", max_e);}    int main () {freopen ("In.txt", "R", stdin);    Freopen ("OUT1.txt", "w", stdout);    int n,m; while (scanf ("%d%d", &n,&m)!=eof) {for (int i=0;i<m;i++) {scanf ("%d%d%d", &e[i].st,&e[i].        En,&e[i].val);        } sort (e,e+m,cmp);        E[m].val=inf;    Solve (n,m); } return 0;}

　　

Nyoj 1208--Water Problem series —————— "DP"