BZOJ1017: [JSOI2008] Warcraft Map Dotr

Portal

The $f[i][j][k]$ indicates that for the $i$ point, contributions to the parent node $j$ the synthesized equipment, cost the $k$, the maximum gain of the power Value.

Simple $f[i][j][k]$ is difficult to transfer, the main reason is unable to maintain the relationship with other Sons. So for each node then a $g[i][j]$ represents the maximum strength value that the former $i$ son of the current point spends for $k$ can Get.

Then be sure to update $g[][]$ and $g[][]$ to update the $f[][][]$.

The state transition equation for $g[i][j]$ is listed As:

$g [cnt][k]=max \{f[son][tol \times edge_v][j] + g[cnt-1][k-j] \}$

Where $tol$ represents the total amount of equipment to be synthesized at the current Point.

Then update the $f[i][j][k]$ according to $g[i][k]$:

$f [i][j][k]= Max \{g[cnt_{max}][k]+ (tol-j) \times power_i \} $

And then you can have a pleasant transfer.

I see other people's code about $tol$ enumeration is decremented thereby reducing unnecessary memset. Not very understanding, hope God Ben Message to tell me qaq, but direct memset will not time Out.

Bzoj 1017//by cydiater//2016.10.26#include <iostream> #include <cstring> #include <string># Include <algorithm> #include <queue> #include <map> #include <ctime> #include <cmath># Include <cstdlib> #include <cstdio> #include <iomanip> #include <bitset>using namespace std;# Define ll long long#define up (i,j,n) for (int. I=j;i<=n;i++) #define down (i,j,n) for (int i=j;i>=n;i--) #define Cmax (a, B) A=max (a, a,) #define Cmin (a, b) a=min (a, b) const int MAXN=1E4+5;CONST int Oo=1000000001;inline int read () {char Ch=getchar ( ); int x=0,f=1;while (ch> ' 9 ' | | ch< ' 0 ') {if (ch== '-') F=-1;ch=getchar ();} While (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} Return x*f;} int N,m,len=0,link[maxn],f[55][105][2005],power[maxn],cost[maxn],lim[maxn],indu[maxn],g[55][maxn],ans=0;bool Tag [maxn];struct Edge{int y,next,v;} E[maxn];namespace solution{inline void Insert (int x,int y,int v) {e[++len].next=link[x]; link[x]=len;e[len].y=y;e[len].v=v;} void Init () {memset (indu,0,sizeof (indu)); N=read (); M=read (); up (i,1,n) lim[i]=oo;up (i,1,n) {power[i]=read (); char ch;scanf ("%c", &ch); tag[i]= (ch== ' A ')? 1:0;if (tag[i ] {int tmp=read (), while (tmp--) {int y=read (), v=read (), Insert (i,y,v); indu[y]++;}} Else{cost[i]=read (); Lim[i]=read ();}}} void TREEDP (int node) {if (!tag[node]) {cmin (lim[node],m/cost[node]); up (i,0,lim[node]) up (j,0,i) f[node][j][i*cost[ node]]=power[node]* (i-j);} Else{lim[node]=oo;for (int I=link[node];i;i=e[i].next) {TREEDP (e[i].y); cmin (lim[node],lim[e[i].y]/e[i].v); cost[node]+=e[i].v*cost[e[i].y];} Cmin (lim[node],m/cost[node]); up (tol,0,lim[node]) {int cnt=0;memset (g,-10,sizeof (g)); g[0][0]=0;for (int I=link[node ];i;i=e[i].next) {cnt++;up (j,0,m) up (k,0,j) Cmax (g[cnt][j],g[cnt-1][j-k]+f[e[i].y][e[i].v*tol][k]);} Up (j,0,tol) up (k,0,m) Cmax (f[node][j][k],g[cnt][k]+ (tol-j) *power[node]);}}} void output (int Node) {up (j,0,lim[node]) up (k,0,m) Cmax (ans,f[node][j][k]);}} int main () {freopen ("input.in", "r", stdin), using namespace solution;init (); memset (f,-10,sizeof (f)); up (i,1,nIf (indu[i]==0) {treedp (i); output (i);} Cout<<ans<<endl;return 0;}

BZOJ1017: [JSOI2008] Warcraft Map Dotr