01 Backpack + Set Information Olympiad one-pass collocation purchase (Buy)

Buy with Match Time limit: Ms Memory limit: 65536 KB Number of submissions: 268 by number: 117 "Title description" Joe thought the clouds were beautiful and decided to go to the shops in the mountains to buy some clouds. There are n clouds in the store, the clouds are numbered as 1,2,......,n, and each cloud has a value. But the store owner told him that some clouds should be paired to buy, so buying a cloud would have to be bought with a cloud that matches the cloud. But Joe has a limited amount of money, so he wants to buy the more value the better. Input Line 1th n,m,w, means n cloud, M collocation, Joe has W's money. Line 2~n+1, each line of Ci,di represents the price and value of I cloud. Line N+2~n+1+m, each line of Ui,vi, said buy the UI must buy VI, similarly, if buy vi must buy UI. Output A row that represents the maximum value that can be obtained. "Input Sample" 5 3 3, 3 3, 5, 1 1 3 3 2 4 2 "output Example" 1 "Hint" "Data Range" 30% Data Guarantee: n≤100 50% Data Guarantee: n≤1,000;m≤100;w≤1,000 100% Data Guarantee: n≤10,000;0≤m≤5000;w≤10,000 The first task of "algorithmic analysis" is to group these clouds together and then treat the set as a whole, and the 01 backpack is Ouke. "Code Analysis" #include <bits/stdc++.h> using namespace std; #define N 10100 int fa[n],c[n],d[n],v[n],w[n]; int find (int x) {return x==fa[x]?x:fa[x]=find (fa[x]);} int main () {int n,m,w1; scanf ("%d%d%d", &AMP;N,&AMP;M,&AMP;W1); for (int i=1;i<=n;i++) {fa[i]=i; scanf ("%d%d", &c[i],&d[i]); } for (int i=1;i<=m;i++) {int x, y; scanf ("%d%d", &x,&y); int R1=find (x), R2=find (y); if (R1!=R2) {fa[r1]=r2;//is purchased together with the same set}} for (int i=1;i<=n;i++) {int X=find (i ); W[x]+=c[i]; v[x]+=d[i];//the total cost of the same set is calculated} int tot=0; for (int i=1;i<=n;i++) {if (fa[i]==i)//One of the collection root nodes is only one, so the root node is a set of clouds {W[++tot]=w[i]; V[tot]=v[i]; }} int f[n]; memset (F,0,sizeof (f)); for (int i=1;i<=tot;i++)//01 backpack for (int j=w1;j>=w[i];j--) F[j]=max (F[j],f[j-w[i]]+v[i]); printf ("%d", f[w1]); return 0; } This code is a little bit confused, later study: #include <bits/stdc++.h> using namespace std; #define N 10100 int fa[n],c[n],d[n]; int find (int x) { Return X==fa[x]?x:fa[x]=find (Fa[x]); } int main () {int n,m,w; scanf ("%d%d%d", &n,&m,&w); for (int i=1;i<=n;i++) {fa[i]=i; scanf ("%d%d", &c[i],&d[i]); } for (int i=1;i<=m;i++) {int x, y; scanf ("%d%d", &x,&y); int R1=find (x), R2=find (y); if (R1!=R2) {fa[r2]=r1;//Why not F[R1]=R2 c[r1]+=c[r2];//cumulatively calculates the total price of each collection D[R1]+=D[R2]; }} int tot=0; for (int i=1;i<=n;i++) {cout<<fa[i]<< "" <<c[i]<< "" <<d[i]<<endl;; if (fa[i]==i) {c[++tot]=c[i]; D[tot]=d[i]; }} int f[n]; memset (F,0,sizeof (f)); for (int. i=1;i<=tot;i++) for (int j=w;j>=c[i];j--) F[j]=max (F[j],f[j-c[i]]+d[i]); printf ("%d", f[w]); return 0; }