uva 529 - Addition Chains
2015-08-08 14:28
375 查看
此题使用迭代加深搜索。。
注意:1.首先确定最少的步数
2.剪枝:获得当前数t之后。。。看 t * 2(的maxd -d次方)能不能大于n,反之则不用再dfs下去了。
//
// main.cpp
// uva 529 - Addition Chains
//
// Created by XD on 15/8/8.
// Copyright (c) 2015年 XD. All rights reserved.
//
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<vector>
#include <string.h>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <cstdio>
using namespace std ;
int n ;
int maxd ;
int ans[1000] ;
int dfs(int d)
{
if (d == maxd) {
if (ans[d-1]==n) {
return 1 ;
}
return 0 ;
}
for (int i = 0; i < d; i++) {
for (int j = i ; j < d; j++) {
if (ans[i] +ans[j] > ans[d-1] && ans[i] + ans[j] <= n ) {
ans[d] = ans[i] + ans[j] ;
int sum = ans[d] ;
for (int t = d +1; t <=maxd; t++) {
sum *= 2 ;
}
if (sum < n ) {
return 0 ;
}
if (dfs(d + 1)) {
return 1;
}
}
}
}
return 0 ;
}
void solved()
{
int len = (int)log2(n) ;
for (maxd = len ;; maxd++) {
ans[0] =1 ;
if (dfs(1)) {
for (int i = 0; i < maxd-1; i++) {
printf("%d " , ans[i]) ;
}
printf("%d\n" , ans[maxd-1]) ;
return ;
}
}
}
int main(int argc, const char * argv[]) {
while (scanf("%d" ,&n)==1&& n!= 0) {
solved() ;
}
return 0;
}
注意:1.首先确定最少的步数
2.剪枝:获得当前数t之后。。。看 t * 2(的maxd -d次方)能不能大于n,反之则不用再dfs下去了。
//
// main.cpp
// uva 529 - Addition Chains
//
// Created by XD on 15/8/8.
// Copyright (c) 2015年 XD. All rights reserved.
//
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<vector>
#include <string.h>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <cstdio>
using namespace std ;
int n ;
int maxd ;
int ans[1000] ;
int dfs(int d)
{
if (d == maxd) {
if (ans[d-1]==n) {
return 1 ;
}
return 0 ;
}
for (int i = 0; i < d; i++) {
for (int j = i ; j < d; j++) {
if (ans[i] +ans[j] > ans[d-1] && ans[i] + ans[j] <= n ) {
ans[d] = ans[i] + ans[j] ;
int sum = ans[d] ;
for (int t = d +1; t <=maxd; t++) {
sum *= 2 ;
}
if (sum < n ) {
return 0 ;
}
if (dfs(d + 1)) {
return 1;
}
}
}
}
return 0 ;
}
void solved()
{
int len = (int)log2(n) ;
for (maxd = len ;; maxd++) {
ans[0] =1 ;
if (dfs(1)) {
for (int i = 0; i < maxd-1; i++) {
printf("%d " , ans[i]) ;
}
printf("%d\n" , ans[maxd-1]) ;
return ;
}
}
}
int main(int argc, const char * argv[]) {
while (scanf("%d" ,&n)==1&& n!= 0) {
solved() ;
}
return 0;
}
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