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<font color="#FF0000">例2.1</font> 背包(knapsack)问题。
<font color="#FF0000">问题: </font>n个物体(物体i的重量为wi,价值为pi)
背包载荷能力m
<font color="#FF0000">求:</font> 把物体i的Xi部分(1≤i≤n ,0≤Xi≤1)装入背包中,
使背包内所放物体的价值最高 <font face="Arial, Helvetica, sans-serif" color="#FF0000">ΣPiXi</font> 且 <font face="Arial, Helvetica, sans-serif" color="#FF0000">ΣWiXi≤m</font>
例如 n=3, m=20
<i><font face="Arial, Helvetica, sans-serif"> (p1, p2, p3)=(25, 24, 15) (w1, w2, w3)=(18, 15, 10)</font></i>
四种可行解
<font color="#FF0000"> (x1, x2, x3) ∑WiXi ∑PiXi</font>
1 (1/2, 1/3,1/4) 16.5 24.25
2 (1, 1/15, 0) 20.0 28.2
3 (0, 2/3, 1) 20.0 31.0
4 (0, 1,1/2) 20.0 31.5
<font color="#FF0000">思路:</font>把单位价值最高的先放入,物2,为24/15。放满15,余5
把次价值再放入,物3为15/10 放10/2 = 5。所以此解最优。</font></b>
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