一道裸的混合背包題目,但是忘記了去重一直TLE,就是如果體積<=完全背包的01,和多重背包都要被完全背包取代,因為他的數量沒限制所以用起來更方便。
#include#include #include #include using namespace std; const int maxn = 210; const int maxc = 200010; inline int read() { char ch; bool flag = false; int a = 0; while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-'))); if(ch != '-') { a *= 10; a += ch - '0'; } else { flag = true; } while(((ch = getchar()) >= '0') && (ch <= '9')) { a *= 10; a += ch - '0'; } if(flag) { a = -a; } return a; } void write(int a) { if(a < 0) { putchar('-'); a = -a; } if(a >= 10) { write(a / 10); } putchar(a % 10 + '0'); } int f[maxc], v[maxn], w[maxn], m[maxn]; int n, V; bool vis[maxn]; void complete_pack(int cost, int weight) { for(int j = cost; j <= V; j++) f[j] = max(f[j], f[j-cost]+weight); } void zero_pack(int cost, int weight) { for(int j = V; j >= cost; j--) f[j] = max(f[j], f[j-cost]+weight); } void multi_pack(int cost, int weight, int amount) { if(cost*amount >= V) { complete_pack(cost, weight); return; } int k = 1; while(k < amount) { zero_pack(cost*k, weight*k); amount -= k; k <<= 1; } zero_pack(amount*cost, amount*weight); } int main() { n = read(); V = read(); memset(vis, false, sizeof(vis)); for(int i=1; i<=n; i++) { v[i] = read(); w[i] = read(); m[i] = read(); } for(int i = 1; i <= n; i++) { for(int j = i+1; j <= n; j++) { if(m[i] == 1) { if(m[j] == -1)if(v[i] >= v[j] && w[i] <= w[j]) vis[i] = true; if(m[j] > 1)if(v[i] >= v[j] && w[i] <= w[j]) vis[i] = true; } else if(m[i] > 1) { if(m[j] == -1) if(v[i] >= v[j] && w[i] <= w[j]) vis[i] = true; } } } for(int i=1; i<=n; i++) { if(m[i]==1 && !vis[i]) zero_pack(v[i], w[i]); else if(m[i]==-1) complete_pack(v[i], w[i]); else if(!vis[i])multi_pack(v[i], w[i],m[i]); } write(f[V]); return 0; }