Bzoj 4004: [JLOI2015] Equipment purchase quasi-array && high extinction

4004: [JLOI2015] Equipment purchase time limit:20 Sec Memory limit:128 MB
submit:337 solved:139
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Face brother recently playing a magical game, this game has n pieces of equipment, each equipped with m properties, with Vector zi (AJ,....., am) expressed (1 <= i <= n; 1 <= J <= m), each equipment needs to spend CI, now face brother want to buy some clothes , but the elder brother is very poor, so always figuring out how to spend as little as possible to buy as much equipment. For the elder brother, if the property of a piece of equipment can be combined with the other equipment purchased (that is, the elder brother can use the equipment on the hand to assemble the effect of the equipment), then this equipment will not buy the necessary. Strict definition is, if the face brother bought Zi1,..... zip this P-piece equipment, then for any pending decision EN, there is no B1,...., bp make B1zi1 + ... + bpzip = zh (b is real), then face brother will buy zh, otherwise zh Facebook brother is useless , nature does not have to buy. For example, Z1 = (1; 2; 3); z2 = (3; 4; 5); zh = (2; 3; 4), B1 =1/2,b2 =1/2, there is b1z1 + b2z2 = zh, then if the face brother bought Z1 and Z2 will not buy zh. Brother-in-the-face wants to spend the least amount of money in the case of buying the most equipment, can you help him to calculate it?

Input

The first line is two numbers n;m.

The next n rows, the number of M per line, where line I describes the property values of the equipment I. The next line is the number of N, where CI represents the cost of purchasing item I equipment. Output

A row of two numbers, the first number indicates the maximum number of items that can be purchased, and the second number represents the maximum number of installed

The minimum cost of the prepared case. Sample Input3 3
1 2 3
3 4 5
2 3 4
1 1 2Sample Output2 2HINT

As described in the title, choose Equipment 1 Equipment 2, equipment 1 equipment 3, equipment 2 equipment 3 are available, but choose equipment 1 and

The cost of equipping 2 is minimal, at 2.

For 100% data, 1 <= n;m <= 500; 0 <= AJ <= 1000.

@idy002 No EPS can still be a la la La.

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cassert>using namespacestd;#defineMAXN 510#defineMOD 1000000007typedefLong LongQword;structitem{intA[MAXN]; intv;} LST[MAXN];BOOLCmp_lst (Constitem& L1,Constitem&L2) {        returnl1.v<L2.V;} Qword MAT[MAXN][MAXN];inttotm=0; Qword pow_mod (Qword X,qword y) {qword ret=1;  while(y) {if(y&1) ret=ret*x%MOD; X=x*x%MOD; Y>>=1; }        returnret;}intMain () {Freopen ("Input.txt","R", stdin); intn,m,x,y,z; scanf ("%d%d",&n,&m);  for(intI=0; i<n;i++)                 for(intj=0; j<m;j++) scanf ("%d",&Lst[i].a[j]);  for(intI=0; i<n;i++) scanf ("%d",&lst[i].v); Sort (Lst,lst+n,cmp_lst); intans1=0, ans2=0;  for(intI=0; i<n;i++)        {                 for(intj=0; j<m;j++) Mat[totm][j]=Lst[i].a[j];  for(intj=0; j<totm;j++) {x=0;  for(intk=0; k<m;k++) {x=K; if(Mat[j][k]) Break; } Qword T=mat[totm][x]/Mat[j][x];  for(intk=0; k<m;k++) Mat[totm][k]= (mat[totm][k]-mat[j][k]*t)%MOD; }                BOOLflag=false;  for(intj=0; j<m;j++) {flag|=Mat[totm][j]; if(flag) Break; }                if(!flag)Continue; Ans1++; Ans2+=lst[i].v; X=0;  for(intj=0; j<m;j++) {x=J; if(Mat[totm][j]) Break; } Qword T=pow_mod (mat[totm][x],mod-2);  for(intj=0; j<m;j++) Mat[totm][j]=mat[totm][j]*t%MOD;  for(intJ=totm;j>0; j--)                {                        BOOLFlag;  for(intk=0; k<m;k++)                        {                                if(mat[j-1][k] | |Mat[j][k]) {Assert (! (mat[j-1][K] &&mat[j][k])); if(mat[j-1][k]) {flag=false; }Else{flag=true; }                                         Break; }                        }                        if(!flag) Break;  for(intk=0; k<m;k++) Swap (Mat[j-1][k],mat[j][k]); } TOTM++; } printf ("%d%d\n", ans1,ans2);}

Bzoj 4004: [JLOI2015] Equipment purchase quasi-array && high extinction