168 条题解
-
0linguist LV 7 @ 2016-07-26 21:15:34
Free Pascal Compiler version 3.0.0 [2015/11/16] for i386
Copyright (c) 1993-2015 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
foo.pas(2,14) Note: Local variable "j" not used
Linking foo.exe
21 lines compiled, 0.0 sec, 27920 bytes code, 1268 bytes data
1 note(s) issued
测试数据 #0: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 804 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 800 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 800 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 804 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 800 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 800 KiB, score = 10
Accepted, time = 45 ms, mem = 804 KiB, score = 100
代码
var
vmax,mmax,i,j,n:integer;
v,m,k:array[1..50] of integer;
function max(x,y:integer):integer;
begin
if x>y then max:=x
else max:=y;
end;
function dp(i,vleft,mleft:integer):integer;
begin
if i>n then dp:=0
else
if (((vleft-v[i])>=0) and ((mleft-m[i])>=0) )then
dp:=max(dp(i+1,vleft-v[i],mleft-m[i])+k[i],dp(i+1,vleft,mleft))
else dp:=dp(i+1,vleft,mleft);
end;
begin
readln(vmax,mmax);
readln(n);
for i:=1 to n do read(v[i],m[i],k[i]);
writeln(dp(1,vmax,mmax));
end.
感觉数据好水,递归居然也可以过0 0 -
02016-07-23 14:37:40@
- 好久没爽爽地一次AC了
- 01背包就是爽
- 前面记忆化搜索的够了
program NASA; var n,V,M,i,j,k:longint; a,b,c:array[1..50]of longint; f:array[1..10000,1..10000]of longint; function max(a,b:longint):longint; begin if a>=b then exit(a) else exit(b); end; begin readln(V,M); readln(n); for i:=1 to n do read(a[i],b[i],c[i]); fillchar(f,sizeof(f),0); for i:=1 to n do for j:=V downto a[i] do for k:=M downto b[i] do f[j,k]:=max(f[j,k],f[j-a[i],k-b[i]]+c[i]); writeln(f[V,M]); end.
-
02016-05-23 23:29:53@
#include <cstdio>
#include <iostream>int main(void){
freopen("in.txt","r",stdin);
int maxv,maxm,n;
int m[60],v[60],e[60];
int f[500][500]={0};
scanf("%d%d%d",&maxv,&maxm,&n);
for(int i=1;i<=n;i++)
scanf("%d%d%d",&v[i],&m[i],&e[i]);
for(int i=1;i<=n;i++)
for(int j=maxv;j>=v[i];j--)
for(int k=maxm;k>=m[i];k--)
f[j][k]=std::max(f[j][k],f[j-v[i]][k-m[i]]+e[i]);
printf("%d",f[maxv][maxm]);
return 0;
} -
02016-04-16 23:46:24@
一开始数组开小了,**我的锅!**
测试数据 #0: Accepted, time = 140 ms, mem = 255072 KiB, score = 10
测试数据 #1: Accepted, time = 203 ms, mem = 255072 KiB, score = 10
测试数据 #2: Accepted, time = 203 ms, mem = 255076 KiB, score = 10
测试数据 #3: Accepted, time = 203 ms, mem = 255072 KiB, score = 10
测试数据 #4: Accepted, time = 203 ms, mem = 255072 KiB, score = 10
测试数据 #5: Accepted, time = 203 ms, mem = 255072 KiB, score = 10
测试数据 #6: Accepted, time = 187 ms, mem = 255072 KiB, score = 10
测试数据 #7: Accepted, time = 218 ms, mem = 255072 KiB, score = 10
测试数据 #8: Accepted, time = 203 ms, mem = 255076 KiB, score = 10
测试数据 #9: Accepted, time = 218 ms, mem = 255076 KiB, score = 10
Accepted, time = 1981 ms, mem = 255076 KiB, score = 100
---------------------------晒一个简单易懂的程序---------------------------
```pascal
type int=longint;
var
v,m,t,i:int;
a,b,d:array[0..401]of int;
c:array[0..401,0..401,0..401]of int;
function max(x,y:int):int;
begin
if x>y then exit(x) else exit(y);
end;function f(x,y,z:int):int;
begin
if c[x,y,z]>0 then exit(c[x,y,z]);
if x=0 then exit(0);
if (a[x]>y)or(b[x]>z) then exit(f(x-1,y,z));
f:=max(f(x-1,y-a[x],z-b[x])+d[x],f(x-1,y,z));
c[x,y,z]:=f;
end;begin
readln(v,m);
readln(t);
for i:=1 to t do readln(a[i],b[i],d[i]);
fillchar(c,sizeof(c),0);
writeln(f(t,v,m));
end.
```
* 我在秋名山等你!!! * -
02016-02-17 18:58:01@
简单的记忆化搜索
c++
#include<stdio.h>
#include<string.h>
int f[51][401][401],V,W;
int v[51],w[51],k[51],n;
inline void max(int& a,int b){if(a<b)a=b;}
int dp(int i,int j,int n){
if(i&&j&&n>=0){
if(~f[n][i][j]) return f[n][i][j];
f[n][i][j]=dp(i,j,n-1);
if(i>=v[n]&&j>=w[n])
max(f[n][i][j],dp(i-v[n],j-w[n],n-1)+k[n]);
return f[n][i][j];
}
return 0;
}
int main(){
memset(f,-1,sizeof f);
scanf("%d%d",&V,&W);
scanf("%d",&n);
for(int i=0;i<n;++i)
scanf("%d%d%d",v+i,w+i,k+i);
printf("%d",dp(V,W,n));
}
-
02016-01-26 15:34:03@
program kkk;
var
i,j,z,vv,mm,n,ii,jj:longint;
l:array[0..10000,0..10000] of longint;
v,m,ka:array[0..100000] of longint;
function maxx(a,b:longint):longint;
begin
if a>b then maxx:=a
else maxx:=b;
end;
begin
readln(vv,mm);
readln(n);
for i:=1 to n do
readln(v[i],m[i],ka[i]);
for i:=1 to n do
begin
for z:=mm downto m[i] do
for j:=vv downto v[i] do
begin
l[j,z]:=maxx(l[j,z],l[j-v[i],z-m[i]]+ka[i]);
end;
end;
write(l[vv,mm]);
end. -
02015-12-15 18:58:39@
#include <iostream>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int N = 405;
int maxv,maxw,n,v[N],w[N],ka[N],f[N][N];
int main(){
scanf("%d%d%d",&maxv,&maxw,&n);
for(int i = 1;i <= n;i ++){
scanf("%d%d%d",&v[i],&w[i],&ka[i]);
}
for(int i = 1;i <= n;i ++){
for(int j = maxv;j >= 0;j --){
for(int k = maxw;k >= 0;k --){
f[j][k] = w[i] <= k &&v[i] <= j ? max(f[j - v[i]][k - w[i]]+ ka[i],f[j][k]) : f[j][k];
}
}
}
printf("%d",f[maxv][maxw]);
return 0;
} -
02015-10-22 15:23:30@
#include <cstdio>
#include <cstring>using namespace std;
inline int max(int a,int b) {
return a>b?a:b;
}int main(void) {
int n=0,v1=0,v2=0;
scanf("%d %d %d",&v1,&v2,&n);
int w[n],c1[n],c2[n];
for (int i=0;i<n;++i)
scanf("%d %d %d",c1+i,c2+i,w+i);
int f[v1+1][v2+1];
memset(f,0,(v1+1)*(v2+1)*sizeof(int));
for (int i=0;i<n;++i)
for (int j=v1;j>=0;--j)
for (int k=v2;k>=0;--k)
if (j>=c1[i] && k>=c2[i])
f[j][k]=max(f[j][k],f[j-c1[i]][k-c2[i]]+w[i]);
int ans=0;
for (int i=0;i<v1+1;++i)
for (int j=0;j<v2+1;++j)
if (f[i][j]>ans)
ans=f[i][j];
printf("%d\n",ans);
return 0;
}QwQ我好菜
-
02015-10-21 19:19:11@
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
int i,j,z,n,f[401][401],w[100],c[100],v[100],W,V,t1,t2;
int main()
{
memset(f,0,sizeof(f));
//freopen("xx.in","r",stdin);
//freopen("xx.out","w",stdout);
cin>>V>>W>>n;
for(i=1;i<=n;++i)
cin>>v[i]>>w[i]>>c[i];
for(i=1;i<=n;++i)
for(j=V;j>=v[i];--j)
for(z=W;z>=w[i];--z)
f[j][z]=max(f[j][z],f[j-v[i]][z-w[i]]+c[i]);
cout<<f[V][W];
} -
02015-10-11 20:07:22@
在经过暴风雨之后,在经过枪林弹雨之后,AC了!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!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-
02015-10-11 20:06:06@
AC了!!!!
-
02015-10-11 20:05:57@
var a,b,s:array[0..50]of longint;
f:array[0..400,0..400]of longint;
v,n,m,i,j,k:longint;
begin
readln(v,m);
readln(n);
for i:=1 to n do
readln(a[i],b[i],s[i]);
for i:=1 to n do
for j:=v downto a[i] do
for k:=m downto b[i] do
if f[j-a[i],k-b[i]]+s[i]>f[j,k] then f[j,k]:=f[j-a[i],k-b[i]]+s[i];
writeln(f[v,m]);
end.嘻嘻,好有成就感!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
-
02015-08-21 19:00:47@
记录信息
评测状态 Accepted
题目 P1334 NASA的食物计划
递交时间 2015-08-21 19:00:10
代码语言 C++
评测机 Jtwd2
消耗时间 45 ms
消耗内存 1136 KiB
评测时间 2015-08-21 19:00:11
评测结果
编译成功测试数据 #0: Accepted, time = 0 ms, mem = 1128 KiB, score = 10
测试数据 #1: Accepted, time = 15 ms, mem = 1132 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 1136 KiB, score = 10
测试数据 #3: Accepted, time = 15 ms, mem = 1136 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 1132 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 1136 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 1132 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 1136 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 1136 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 1136 KiB, score = 10
Accepted, time = 45 ms, mem = 1136 KiB, score = 100
代码
#include <iostream>
#include <stdio.h>
using namespace std;
int f[410][410];
int wei[51];
int big[51];
int krl[51];
int main()
{
int n;
int r,p;
scanf("%d%d",&r,&p);
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d%d%d",&big[i],&wei[i],&krl[i]);for(int i=1;i<=n;i++)
for(int j=r;j>=big[i];j--)
for(int k=p;k>=wei[i];k--)
{
f[j][k]=max(f[j-big[i]][k-wei[i]]+krl[i],f[j][k]);
}
int maxans=-1000000000;
for(int j=r;j;j--)
for(int k=p;k;k--)
{
maxans=max(maxans,f[j][k]);
}
printf("%d",maxans);
} -
02015-02-03 14:50:39@
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <cstdlib>
using namespace std;
const int MAXN=51,MAXV=401,MAXL=401;
int f[MAXN][MAXV][MAXL],a[MAXN],b[MAXN],c[MAXN];
int main(){
int i,j,k,n,m,v,g;
cin>>v>>g;
cin>>n;
for(i=1;i<=n;i++) cin>>a[i]>>b[i]>>c[i];
for(i=1;i<=n;i++) f[i][0][0]=0;
for(j=0;j<=v;j++) f[0][j][0]=0;
for(k=0;k<=g;k++) f[0][0][k]=0;
for(i=1;i<=n;i++)
for(j=0;j<=v;j++)
for(k=0;k<=g;k++){
f[i][j][k]=f[i-1][j][k];
if(j>=a[i] && k>=b[i]){
f[i][j][k]=max(f[i-1][j][k],f[i-1][j-a[i]][k-b[i]]+c[i]);
}
}
cout<<f[n][v][g]<<endl;
system("pause");
return 0;
} -
02015-01-22 16:27:29@
#include <iostream>
#include <algorithm>
#define MAXL 401
using namespace std;int dp[MAXL][MAXL], W, V, n, w, v, p;
void ZeroOnePack(int v, int w, int p);
int main()
{
cin >> V >> W;
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> v >> w >> p;
ZeroOnePack(v, w, p);
}
cout << dp[V][W] << endl;
return 0;
}void ZeroOnePack(int v, int w, int p)
{
for (int i = V; i >= v; i--)
for (int j = W; j >= w; j--)
dp[i][j] = max(dp[i - v][j - w] + p, dp[i][j]);
}
N个地方把v和w整反了,错了N次。。。 -
02015-01-15 17:50:11@
感觉动态规划就是枚举嘛,诶,就是把所有状态枚举一遍,比纯枚举排除了很多情况罢了- -所以二维费用比一维费用差别就在于枚举的东西更多了而已。。吧。。
###blcok code
program P1334;
uses math;
var data:array[0..400,0..400] of longint;
i,v,a,b,n,maxa,maxb,j,k:longint;
begin
fillchar(data,sizeof(data),0); read(maxa,maxb);
read(n);
for i:=1 to n do
begin
read(a,b,v);
for j:=maxa downto a do
for k:=maxb downto b do
data[j,k]:=max(data[j,k],data[j-a,k-b]+v);
end;write(data[maxa,maxb]);
end. -
02014-10-26 10:54:00@
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;struct food{int v,m,c;}g[55];
int n,m,v,f[405][405];int main()
{
scanf("%d%d%d",&v,&m,&n);
for(int i=1;i<=n;i++)
scanf("%d%d%d",&g[i].v,&g[i].m,&g[i].c);
for(int i=1;i<=n;i++)
for(int j=v;j>=g[i].v;j--)
for(int k=m;k>=g[i].m;k--)
f[j][k]=max(f[j][k],f[j-g[i].v][k-g[i].m]+g[i].c);
printf("%d\n",f[v][m]);
return 0;
} -
02014-08-21 16:58:57@
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
int V, W, N, f[410][410], vv[55], ww[55], cc[55];
int main()
{
scanf("%d%d%d", &V, &W, &N);for(int i = 1; i <= N; i++) cin >> vv[i] >> ww[i] >> cc[i];
for(int i = 1; i <= N; i++)
for(int j = V; j >= vv[i]; j--)
for(int k = W; k >= ww[i]; k--)
if(f[j - vv[i]][k - ww[i]] + cc[i] > f[j][k])
f[j][k] = f[j - vv[i]][k - ww[i]] + cc[i];cout << f[V][W] << endl;
return 0;
} -
02013-11-08 09:04:45@
var
a,b,s:array[0..50] of longint;
f:array[0..400,0..400] of longint;
n,v,m,i,j,k:longint;
begin
readln(v,m);
readln(n);
for i:=1 to n do
readln(a[i],b[i],s[i]);
for i:=1 to n do
for j:=v downto a[i] do
for k:=m downto b[i] do
if f[j-a[i],k-b[i]]+s[i]>f[j,k] then f[j,k]:=f[j-a[i],k-b[i]]+s[i];
writeln(f[v,m]);
end.
现在每做一道题,写一份题解,
但这个太水了
不写了 -
02013-10-15 15:15:31@
#include <iostream>
//#include <memory.h>using namespace std;
const int MAX_SIZE = 1000;
int v[MAX_SIZE] = {0}, m[MAX_SIZE] = {0}, c[MAX_SIZE] = {0}, Calorie[MAX_SIZE][MAX_SIZE];
int main()
{
//
// 清零操作(可忽略)
//
//memset(Calorie, 0, MAX_SIZE * MAX_SIZE * sizeof(int));//
// 输入数据
//
int volumeMax, weightMax, N;
cin >> volumeMax >> weightMax;
cin >> N;//
// 读入每个食物的体积、质量和所含卡路里
//
int i, j, k;
for (i = 1; i <= N; i++) {
cin >> v[i] >> m[i] >> c[i];
}//
// 背包DP
//
for (i = 1; i <= N; i++) {
for (j = volumeMax; j >= v[i]; j--) {
for (k = weightMax; k >= m[i]; k--) {
if (Calorie[j - v[i]][k - m[i]] + c[i] > Calorie[j][k]) {
Calorie[j][k] = Calorie[j - v[i]][k - m[i]] + c[i];
}
}
}
}//
// 输出结果
//
cout << Calorie[volumeMax][weightMax];return 0;
}