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https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
53 lines
1.2 KiB
C++
53 lines
1.2 KiB
C++
/*原题是这样的：
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对于20%数据,1<=n<=10;
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对于100%数据,1<=rmb<=100,1<=rp<=100,1<=time<=1000;
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对于100%数据,1<=m<=100,1<=r<=100,1<=n<=100.*/
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#include <cstdio>
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#include <cstdlib>
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#include <cstring>
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#include <algorithm>
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using namespace std;
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#define MAXN 101
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#define MAXV 101
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#define MAXU 101
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/** f[x][y] means GG can find f[x][y] MM by x of V and y of U*/
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int f[MAXV][MAXU];
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/** time[x][y] means it takes time[x][y] while x of V and y of U is taken,*/
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int time[MAXV][MAXU];
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int v[MAXN];
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int u[MAXN];
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int w[MAXN];
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int main()
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{
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int n;
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scanf("%d",&n);
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for(int i=0;i<n;i++)
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{
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scanf("%d %d %d",&v[i],&u[i],&w[i]);
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}
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int m,r;
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scanf("%d %d",&m,&r);
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for(int i=0;i<n;i++)
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{
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for(int j=m;j>=v[i];j--)
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{
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for(int k=r;k>=u[i];k--)
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{
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if(f[j][k]<f[j-v[i]][k-u[i]]+1)
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{
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f[j][k]=f[j-v[i]][k-u[i]]+1;
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time[j][k]=time[j-v[i]][k-u[i]]+w[i];
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}
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else if(f[j][k]==f[j-v[i]][k-u[i]]+1)
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{
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time[j][k]=min(time[j][k],time[j-v[i]][k-u[i]]+w[i]);
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}
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}
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}
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}
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printf("%d\n",time[m][r]);
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return 0;
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}
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