2.3.1 记忆化搜索与动态规划

1. 01背包问题

<pre><code>
//
// Created by Nathan on 15/3/23.
// Copyright (c) 2015年 Nathan. All rights reserved.
//
// 1<=n<=100;
// 1<=wi,vi<=100;
// 1<=W<=10000;
// n=4;
// (w,v) ={(2,3),(1,2),(3,4),(2,2)}
// W=5;
//

include <iostream>

using namespace std;
int n,W;
const int MAX = 100;
int w[MAX],v[MAX];
int solve(){
int MAX_N = 100, MAX_W = 10000;
int dp[MAX_N+1][MAX_W+1];
for (int i =0 ; i<n; i++) {
for (int j=0; j<=W; j++) {
if(j>=w[i]){
dp[i+1][j] = max( dp[i][j],dp[i][j-w[i]]+v[i]);
} else {
dp[i+1][j] = dp[i][j];
}
}
}
return dp[n][W];
}
int solve_single(){
int MAX_W = 10000;
int dp[MAX_W];
for (int i=0; i<n; i++) {
for (int j = W; j >= 0; j--) {
if( j>=w[i] ){
dp[j] = max(dp[j], dp[j-w[i]]+v[i] );
}
}
}
return dp[W];
}
int main(int argc, const char * argv[]) {
cin >> n >> W;
for (int i = 0; i < n; i++) {
cin >> w[i];
}
for (int i = 0; i < n; i++) {
cin >> v[i];
}
cout << "Solve_singe:" << solve_single() <<endl;
cout << "Solve:" << solve() <<endl;
return 0;
}
</code></pre>

2. 最长公共子序列问题

<pre><code>
//
// Created by Nathan on 15/3/25.
// Copyright (c) 2015年 Nathan. All rights reserved.
//
// 1 <=n,m <=1000
// n = 4;
// m = 4;
// s = 'abcd';
// t = 'becd';
//

include <iostream>

using namespace std;
const int MAX = 1000;
int dp[MAX+1][MAX+1];
int n,m;
char s[MAX],t[MAX];
int solve(){
for (int i = 0; i<n; i++) {
for (int j = 0; j<m; j++) {
if(s[i]==t[j]){
dp[i+1][j+1] = dp[i][j]+1;
} else {
dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);
}
}
}
return dp[n][m];
}
int main(int argc, const char * argv[]) {
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> s[i];
}
for (int i = 0; i < m; i++) {
cin >> t[i];
}
cout << solve() << endl;
return 0;
}
</code></pre>

2.3.2 进一步探讨递推关系

1. 完全背包问题1

<pre><code>
//
// Created by Nathan on 15/3/27.
// Copyright (c) 2015年 Nathan. All rights reserved.
// 完全背包1
// 每件物品可以挑选任意多件
// 1 <= n <= 100
// 1 <= wi,vi <= 100
// 1 <= W <= 10000
// n = 3
// (w,v) = {(3,4),(4,5),(2,3)}
// W = 7
//

include <iostream>

using namespace std;
const int MAX = 100;
const int MAX_R = 10000;
int n,W;
int w[MAX],v[MAX];
int solve_single(){
int dp[MAX_R];
for (int i = 0; i<n; i++) {
for (int j=w[i]; j<=W; j++) {
dp[j] = max(dp[j],dp[j-w[i]]+v[i] );
}
}
return dp[W];
}
int solve(){
int dp[MAX+1][MAX_R+1];
for (int i = 0; i<n; i++) {
for (int j = 0; j<=W; j++) {
if( j>=w[i] ){
dp[i+1][j] = max(dp[i][j], dp[i+1][j-w[i]]+v[i]);
} else {
dp[i+1][j] = dp[i][j];
}
}
}
return dp[n][W];
}
int main(int argc, const char * argv[]) {
cin >> n >> W;
for (int i =0; i<n; i++) {
cin >> w[i];
}
for (int i =0; i<n; i++) {
cin >> v[i];
}
cout << "Solve:" << solve() << endl;
cout << "Solve_single:" << solve_single() << endl;
return 0;
}
</code></pre>

2. 完全背包问题2

<pre><code>
//
// main.cpp
// 2.3.2.2 Package_150328
//
// Created by Nathan on 15/3/28.
// Copyright (c) 2015年 Nathan. All rights reserved.
// 1<= n <= 100;
// 1<= wi <= 10000000;
// 1<= vi <= 100;
// 1<= W <= 1000000000;
// n = 4
// (w,v) = {(2,3),(1,2),(3,4),(2,2)}
// W = 5
//

include <iostream>

using namespace std;
int n,W;
const int MAX_N = 100;
const int MAX_V = 100;
const int MAX_W = 10000000;
const int INF = 1000000;
int w[MAX_W],v[MAX_V],dp[MAX_N+1][MAX_NMAX_V+1];
int solve(){
fill(dp[0], dp[0]+MAX_NMAX_V+1, INF);
dp[0][0]=0;
for (int i = 0; i<n; i++) {
for (int j = 0 ; j<=MAX_NMAX_V; j++) {
if(j<v[i]){
dp[i+1][j] = dp[i][j];
} else {
dp[i+1][j] = min(dp[i][j], dp[i][j-v[i]]+w[i]);
}
}
}
int res = 0;
for (int i =0 ; i<=MAX_NMAX_V; i++) {
if(dp[n][i]<=W){
res = i;
}
}
return res;
}
int main(int argc, const char * argv[]) {
cin >> n >> W;
for (int i = 0; i < n; i++) {
cin >> w[i];
}
for (int i = 0; i < n; i++) {
cin >> v[i];
}
cout << solve() << endl;
return 0;
}
</code></pre>

3. 多重部分和问题

<pre><code>
//
// main.cpp
// 2.3.2.3 Multi-section_150328
//
// Created by Nathan on 15/3/28.
// Copyright (c) 2015年 Nathan. All rights reserved.
// 3 17
// 3 5 8
// 3 2 2
//

include <iostream>

using namespace std;
int n,K;
const int MAX_N = 100;
const int MAX_K = 100000;
int a[MAX_N],m[MAX_N];
int dp[MAX_N+1][MAX_K+1];
int solve(){
dp[0][0]=true;
for (int i=0; i<n; i++) {
for (int j=0; j<=K; j++) {
for (int k =0; ka[i]<=j&&k<=m[i]; k++) {
dp[i+1][j] |= dp[i][j-ka[i]];
}
}
}
for (int i=0; i<=n; i++) {
for (int j=0; j<=K; j++) {
cout << dp[i][j] << " ";
}
cout << endl;
}
return dp[n][K];
}
int main(int argc, const char * argv[]) {
cin >> n >> K;
for (int i = 0; i<n; i++) {
cin >> a[i];
}
for (int i = 0; i<n; i++) {
cin >> m[i];
}
int res = solve();
cout << res <<endl;
return 0;
}
</code></pre>

4.最长上升子序列

<pre><code>
//
// main.cpp
// 2.3.2.4 LIS_150328
//
// Created by Nathan on 15/3/28.
// Copyright (c) 2015年 Nathan. All rights reserved.
//

include <iostream>

using namespace std;
const int MAX = 1000;
int n,a[1000000];
int dp[MAX];
const int INF = 100000000;
int solve_sec(){
fill(dp, dp+n, INF);
for (int i = 0; i<n; i++) {
*lower_bound(dp, dp+n, a[i])=a[i];
}
cout << lower_bound(dp, dp+n, INF)-dp;
return 0;
}
int solve(){
int res = 0;
for (int i =0; i<n; i++) {
dp[i]=1;
for (int j=0; j<i; j++) {
if (a[j]<a[i]) {
dp[i]=max( dp[i], dp[j]+1);
}
}
res = max( res , dp[i]);
for (int k=0; k<n; k++) {
cout << dp[k] << " " ;
}
cout << endl;
}
return res;
}
int main(int argc, const char * argv[]) {
cin >> n;
for (int i = 0; i<n; i++) {
cin >> a[i];
}
//solve_sec();
//int res = solve_sec();
cout << solve() << endl;
return 0;
}
</code></pre>

5. Lower_bound

lower_bound是STL(标准模板库)中的一个函数。
<code>lower_bound(<#_ForwardIterator __first#>, <#_ForwardIterator __last#>, <#const _Tp &_value#>)</code>
该函数的作用是从已拍好的序列a中利用二分搜索找出指向满足ai≥k的ai指针。类似的upper_bound是找出满足ai>k的ai的最小指针。
<pre><code>

include <iostream>

include <algorithm>

include <vector>

using namespace std;
int main () {
int myints[] = {10,20,30,30,20,10,10,20};
vector<int> v(myints,myints+8); // 10 20 30 30 20 10 10 20
sort (v.begin(), v.end()); // 10 10 10 20 20 20 30 30
vector<int>::iterator low,up;
low=lower_bound (v.begin(), v.end(), 10); // ^
up= upper_bound (v.begin(), v.end(), 10); // ^
cout << "lower_bound at position " << (low- v.begin()) << endl;
cout << "upper_bound at position " << (up - v.begin()) << endl;
return 0;
}
</code></pre>