1.AcWing 2.01背包问题
分析思路
![C++算法之数学与简单DP_第1张图片](http://img.e-com-net.com/image/info8/a6ae1b1bfa8f489aa04614b743e7aa4e.jpg)
代码实现
#include
using namespace std;
const int N=1010;
int dp[N][N];
int v[N],w[N];
int n,m;
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++) cin>>v[i]>>w[i];
for(int i=1;i<=n;i++)
{
for(int j=0;j<=m;j++)
{
dp[i][j]=dp[i-1][j];//左边
if(j>=v[i]) dp[i][j]=max(dp[i][j],dp[i-1][j-v[i]]+w[i]);
}
}
cout<
2.AcWing 1015.摘花生
分析思路
![C++算法之数学与简单DP_第2张图片](http://img.e-com-net.com/image/info8/e486311794074ff1935ddbbacb406f12.jpg)
代码实现
#include
#include
using namespace std;
const int N =110;
int f[N][N];
int w[N][N];
int main()
{
int n;
cin>>n;
while(n--)
{
int r,c;
cin>>r>>c;
for(int i=1;i<=r;i++)
{
for(int j=1;j<=c;j++)
{
cin>>w[i][j];
f[i][j]=max(f[i-1][j]+w[i][j],f[i][j-1]+w[i][j]);
}
}
cout<
3.AcWing 895.最长上升子序列
分析思路
![C++算法之数学与简单DP_第3张图片](http://img.e-com-net.com/image/info8/fd9fec86fa6d44ffa0a4dc05ef238bd9.jpg)
代码实现
#include
#include
#include
using namespace std;
const int N=1010;
int f[N];
int w[N];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++) scanf("%d",&w[i]);
int res=0;
for(int i=1;i<=n;i++)
{
f[i]=1;//只有一个数
for(int j=1;jw[j]) f[i]=max(f[i],f[j]+1);
}
res=max(res,f[i]);
}
cout<
4.AcWing 1212.地宫取宝
分析思路
![C++算法之数学与简单DP_第4张图片](http://img.e-com-net.com/image/info8/0fcf772513f0415e94023d0e7c453da0.jpg)
代码实现
#include
#include
#include
using namespace std;
const int N = 55, MOD = 1000000007;
int n, m, k;
int w[N][N];
int f[N][N][13][14];
int main()
{
cin >> n >> m >> k;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
{
cin >> w[i][j];
w[i][j] ++ ;
}
f[1][1][1][w[1][1]] = 1;
f[1][1][0][0] = 1;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ )
{
if (i == 1 && j == 1) continue;
for (int u = 0; u <= k; u ++ )
for (int v = 0; v <= 13; v ++ )
{
int &val = f[i][j][u][v];
val = (val + f[i - 1][j][u][v]) % MOD;
val = (val + f[i][j - 1][u][v]) % MOD;
if (u > 0 && v == w[i][j])
{
for (int c = 0; c < v; c ++ )
{
val = (val + f[i - 1][j][u - 1][c]) % MOD;
val = (val + f[i][j - 1][u - 1][c]) % MOD;
}
}
}
}
int res = 0;
for (int i = 0; i <= 13; i ++ ) res = (res + f[n][m][k][i]) % MOD;
cout << res << endl;
return 0;
}