总结:
1.
50分的代码(因为由多个数据,所以不能改变num[]数组)
#include<bits/stdc++.h> using namespace std; struct node{ int pos, value; }que[1005]; int c[1005], w[1005], num[1005], f[1005], vis[1005]; int n, head, tail, V, q, h, z[1005]; int main() { cin >> n; for(int i = 0; i < n; i++) cin >> c[i] >> w[i] >> num[i]; cin >> q; for(int cnt = 1; cnt <= q; cnt++) { memset(f, 0, sizeof(f)); cin >> h >> V; vis[h] = 1; for(int i=0;i<n;i++) { if(!vis[i]) { z[i] = min(num[i], V/c[i]); //用z[]数组代表能用多少件物品 for(int mo=0;mo<c[i];mo++) { head=tail=0; for(int k=0;k<=(V-mo)/c[i];k++) { int x=k; int y=f[k*c[i]+mo]-k*w[i]; while(head<tail && que[head].pos<k-z[i])head++; while(head<tail && que[tail-1].value<=y)tail--; que[tail].value=y,que[tail].pos=x; tail++; f[k*c[i]+mo]=que[head].value+k*w[i]; } } } } cout << f[V] << endl; vis[h] = 0; } return 0; }