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装载问题
装载问题同样也是递归回溯法的一个简单应用,用子集树表示解空间显然是最合适的。在递归回溯时,可以进行相应的剪枝。问题的解要满足两个条件:
1.首先将第一艘轮船尽可能装满
2.将剩余的集装箱装上第二艘轮船
由此可知,只要求出不超过第一艘轮船载重量c1的最大值,若总重量-c1<=c2则可以装上两艘船。
/* * @author Panoss */ #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<vector> #include<ctime> #include<stack> #include<queue> #include<list> using namespace std; #define DBG 1 #define fori(i,a,b) for(int i = (a); i < (b); i++) #define forie(i,a,b) for(int i = (a); i <= (b); i++) #define ford(i,a,b) for(int i = (a); i > (b); i++) #define forde(i,a,b) for(int i = (a); i >= (b); i++) #define forls(i,a,b,n) for(int i = (a); i != (b); i = n[i]) #define mset(a,v) memset(a, v, sizeof(a)) #define mcpy(a,b) memcpy(a, b, sizeof(a)) #define dout DBG && cerr << __LINE__ << " >>| " #define checkv(x) dout << #x"=" << (x) << " | "<<endl #define checka(array,a,b) if(DBG) { \ dout<<#array"[] | " << endl; forie(i,a,b) cerr <<"["<<i<<"]="<<array[i]<<" |"<<((i-(a)+1)%5?" ":"\n"); if(((b)-(a)+1)%5) cerr<<endl; } #define redata(T, x) T x; cin >> x #define MIN_LD -2147483648 #define MAX_LD 2147483647 #define MIN_LLD -9223372036854775808 #define MAX_LLD 9223372036854775807 #define MAX_INF 18446744073709551615 inline int reint() { int d; scanf("%d",&d); return d; } inline long relong() { long l; scanf("%ld",&l); return l; } inline char rechar() { scanf(" "); return getchar(); } inline double redouble() { double d; scanf("%lf", &d); return d; } inline string restring() { string s; cin>>s; return s; } int c1, c2, n, W[15]; int current_weight, best_weight, residue_weight, total_weight; void init() { mset(W, 0); current_weight = 0; best_weight = 0; total_weight = 0; } void DFS(int depth) { if(depth > n) { best_weight = max(best_weight, current_weight); return ; } residue_weight -= W[depth]; if(current_weight + W[depth] <= c1) //搜索左子树 { current_weight += W[depth]; DFS(depth + 1); current_weight -= W[depth]; //回溯时一定要改回来,因为状态不改变 } if(current_weight + residue_weight > best_weight) //搜索右子树,注意剪枝 DFS(depth+1); residue_weight += W[depth]; } int main() { //freopen("data.txt","r",stdin); while(scanf("%d%d%d",&c1,&c2,&n)==3&&(c1+c2+n)) { init(); forie(i,1,n) { scanf("%d",&W[i]); total_weight += W[i]; } residue_weight = total_weight; DFS(1); if(total_weight - best_weight <= c2) cout << "Yes" << endl; else cout <<"No" << endl; } return 0; }
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