UVa 990 - Diving for Gold
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題目:一個潛水員冒險家,在水下發現了一個沉船,他只有一個氧氣瓶,可以提供t時間的氧氣;
沉船裡面有一下寶箱,他們的價值和深度已知,取得每個寶物的代價是3*w*d(i),求最大價值。
分析:dp,01背包。數據較少,直接利用位狀態記錄每個元素的選取情況。
說明:直接將t/3w,可以調高計算效率。
#include <cstring>#include <cstdio>int v[30], d[30], f[3334], s[3334];int main(){int t, w, n, cases = 0;while (~scanf("%d%d%d",&t,&w,&n)) {for (int i = 0; i < n; ++ i)scanf("%d%d",&d[i],&v[i]);t /= 3*w;memset(f, 0, sizeof(f));memset(s, 0, sizeof(s));for (int i = 0; i < n; ++ i)for (int j = t; j >= d[i]; -- j)if (f[j] < f[j-d[i]]+v[i]) {f[j] = f[j-d[i]]+v[i];s[j] = s[j-d[i]]|(1<<i);}int sum = 0;for (int i = 0; i < 30; ++ i)if(s[t]&(1<<i))sum ++;if (cases ++) printf("\n");printf("%d\n%d\n",f[t],sum);for (int i = 0; i < 30; ++ i) if (s[t]&(1<<i))printf("%d %d\n",d[i],v[i]);}return 0;}
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