Acwing 动态规划打卡
背包问题
多重背包问题 I
https://www.acwing.com/problem/content/4/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<unordered_map>
using namespace std;
typedef long long LL;
int N,V;
int v[105],w[105],s[105];
int f[105][105];
int main(){
cin>>N>>V;
for(int i=1;i<=N;i++){
cin>>v[i]>>w[i]>>s[i];
}
for(int i=1;i<=N;i++){
for(int j=0;j<=V;j++){
for(int k=0;k<=s[i];k++){
if(j>=k*v[i]){
f[i][j]=max(f[i][j],f[i-1][j-k*v[i]]+k*w[i]);
}
}
}
}
cout<<f[N][V];
return 0;
}
// freopen("testdata.in", "r", stdin);
多重背包问题 II
https://www.acwing.com/problem/content/5/
转化成01背包问题,同时划分背包时不一个个分,而是通过2的次方数拆分
这样也可以表示出1到n的所有数
#include <iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
using namespace std;
int N,V;
struct Good{
int v,w;
};
vector<Good> goods;
int v,w,s;
int dp[2005];
int main(){
cin>>N>>V;
while(N--){
cin>>v>>w>>s;
for(int k=1;k<=s;k*=2){
s-=k;
goods.push_back({k*v,k*w});
}
if(s>0){
goods.push_back({s*v,s*w});
}
}
for(auto g:goods){
for(int j=V;j>=g.v;j--){
dp[j]=max(dp[j],dp[j-g.v]+g.w);
}
}
cout<<dp[V];
return 0;
}
// freopen("testdata.in", "r", stdin);
分组背包问题
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int n, m;
int v[N][N], w[N][N], s[N];
int f[N];
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i ++ )
{
cin >> s[i];
for (int j = 0; j < s[i]; j ++ )
cin >> v[i][j] >> w[i][j];
}
for (int i = 1; i <= n; i ++ )
for (int j = m; j >= 0; j -- )
for (int k = 0; k < s[i]; k ++ )
if (v[i][k] <= j)
f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
cout << f[m] << endl;
return 0;
}
线性DP
898. 数字三角形
https://www.acwing.com/problem/content/900/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<unordered_map>
using namespace std;
const int N=510;
int n;
int w[N][N],f[N][N];
//f(i,j) 从下往上走 到i,j的最大路径
int main(){
cin>>n;
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
cin>>w[i][j];
}
}
for(int i=1;i<=n;i++){
f[n][i]=w[n][i];
}
for(int i=n-1;i;i--){
for(int j=1;j<=n;j++){
f[i][j]=max(f[i+1][j]+w[i][j],f[i+1][j+1]+w[i][j]);
}
}
cout<<f[1][1]<<endl;
return 0;
}
// freopen("testdata.in", "r", stdin);
895. 最长上升子序列
https://www.acwing.com/problem/content/897/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<unordered_map>
using namespace std;
typedef long long LL;
int a[1005];
int f[1005];//f(i)以i为结尾的最长上升子序列
int n;
int main(){
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
}
for(int i=1;i<=n;i++){
f[i]=1;
for(int j=1;j<i;j++){
if(a[j]<a[i]){
f[i]=max(f[i],f[j]+1);
}
}
}
int res=0;
for(int i=1;i<=n;i++){
res=max(res,f[i]);
}
cout<<res<<endl;
return 0;
}
// freopen("testdata.in", "r", stdin);
状态计算找最后一个不同点,由于方程确定了结尾是不变的,我们就去找倒数第二个点。就有0到i个选择。
最长公共子序列
https://www.acwing.com/problem/content/899/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<unordered_map>
using namespace std;
typedef long long LL;
int n,m;
char a[1005],b[1005];
int f[1005][1005];//f(i,j)1到i结尾 1到j的最长公共子序列
int main(){
cin>>n>>m>>a+1>>b+1;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
f[i][j]=max(f[i-1][j],f[i][j-1]);
if(a[i]==b[j]){
f[i][j]=max(f[i][j],f[i-1][j-1]+1);
}
}
}
cout<<f[n][m];
return 0;
}
// freopen("testdata.in", "r", stdin);
最短编辑距离
https://www.acwing.com/problem/content/904/
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e3+5;
char a[N],b[N];
int n,m;
int f[N][N];
int main()
{
cin>>n>>(a+1)>>m>>(b+1);
for(int i=0;i<=n;i++) f[i][0]=i;
for(int j=0;j<=m;j++) f[0][j]=j;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
int temp;
if(a[i]==b[j]) temp=0;
else temp=1;
f[i][j]=min(f[i-1][j]+1,min(f[i][j-1]+1,f[i-1][j-1]+temp));
}
}
cout<<f[n][m]<<endl;
}
编辑距离
https://www.acwing.com/problem/content/901/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<unordered_map>
using namespace std;
typedef long long LL;
int n,m;
char s[1005][20];
int f[1005][1005];
int fun(char a[],char b[]){
int lena=strlen(a+1);
int lenb=strlen(b+1);
for(int i=0;i<=lena;i++) f[i][0]=i;
for(int j=0;j<=lenb;j++) f[0][j]=j;
for(int i=1;i<=lena;i++){
for(int j=1;j<=lenb;j++){
int temp=1;
if(a[i]==b[j]){
temp=0;
}
f[i][j]=min(f[i-1][j]+1,min(f[i][j-1]+1,f[i-1][j-1]+temp));
}
}
return f[lena][lenb];
}
int main(){
cin>>n>>m;
for (int i = 0; i < n; i ++ ) scanf("%s", s[i] + 1);
while(m--){
int ans=0;
int num;
char b[1005];
scanf("%s%d", b + 1, &num);
for(int i=0;i<n;i++){
int temp=fun(s[i],b);
if(temp<=num){
ans++;
}
}
cout<<ans<<endl;
}
return 0;
}
// freopen("testdata.in", "r", stdin);
274. 移动服务
https://www.acwing.com/problem/content/description/276/
计数类DP
900. 整数划分
https://www.acwing.com/problem/content/902/
#include <iostream>
#include <cstring>
#include <algorithm>
const int mod=1e9+7;
using namespace std;
int f[1005],n;
int main()
{
cin>>n;
f[0]=1;
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
f[i]=(f[i]+f[i-j])%mod;
}
}
cout<<f[n]<<endl;
}
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