0-1背包问题——使用动态规划方法解决

思路：向后处理法

fi(x)=max{fi-1(x), fi-1(x-wi)+pi}

 

/*背包问题——使用动态规划方法解决 */ #include <stdio.h>void DKNAP_Pack(float p[], float w[], float M, int n, int x[]){ float P[1000],W[1000],pp,ww; int F[100]; int i,j,k,l,h,next,u; l = h = 1; F[0] = 1; P[1] = W[1] = 0; F[1] = next = 2; for(i=1; i<n; i++) { k = l; u = l; while(u<=h && W[u]+w[i]<M) u++; if(u>h) u--; for(j=l; j<=u; j++) { pp = P[l] + p[i]; ww = W[l] + w[i]; while(k<=h && W[k]<ww) { P[next] = P[k]; W[next] = W[k]; next++; k++; }; if(k<=h && W[k]==ww) { if(P[k] > pp) pp = P[k]; k++; } if(pp > P[next-1]) { P[next] = pp; W[next] = ww; next++; } while(k<=h && P[k]<=P[next-1]) k++; }//for while(k<=h) { P[next] = P[k]; W[next] = W[k]; next++; k++; } l = h+1; h = next-1; F[i+1] = next; }//for /*有了S(n-1)，求最终答案*/ i = F[n-1]; while(i<F[n] && w[n]+W[i]<=M) i++; i--; if(P[i]+p[n] > P[F[n]-1]) { P[F[n]] = P[i]+p[n]; W[F[n]] = W[i]+w[n]; }else { P[F[n]] = P[F[n]-1]; W[F[n]] = W[F[n]-1]; } printf("P=%f, W=%f./n", P[F[n]], W[F[n]]); /*求x[n]，即w的组成成分*/ for(i=0; i<=n; i++) printf("%d ",F[i]); printf("Ouput p and w:/n"); for(i=F[0]; i<=F[n]; i++) printf("%f %f/n",P[i], W[i]); printf("/n"); pp = P[F[n]]; ww = W[F[n]]; for(i=n-1; i>=0; i--) { j=F[i]; x[i+1] = 0; while(j<F[i+1]) { if(pp==P[j] && ww==W[j]) { x[i+1] = 0; break; } if(pp-p[i+1]==P[j] && ww-w[i+1]==W[j]) { x[i+1] = 1; pp = pp-p[i+1]; ww = ww-w[i+1]; break; } j++; }//while }//for }int main(){ int i, n, x[10]; float p[10],w[10],M; M = 6; n = 3; p[1] = 1; p[2] = 2; p[3] = 5; w[1] = 2; w[2] = 3; w[3] = 4; DKNAP_Pack(p, w, M, n, x); for(i=1; i<=n; i++) printf("%d ",x[i]); getchar(); getchar(); getchar(); return 0;} 

 

算法空间复杂度是O(2^n)，但是由于M比2^n小很多，而且根据支配规则（wi>wj时，必须有pi>pj）可以清楚一些不满足条件的序偶，故实际上，算法是可行的。