【DP】【lis】【lcs】【合并石子】【背包问题】DP基础问题四则

Re.DP复习题
1.lis 最长上升序列
输入 第一行一个整数n，第二行n个整数
输出 lis长度

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#define LL long long#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;const int maxn = 10005;int n;int dp[maxn],m[maxn];void lis(){    int maxxx = 0;    for(int i = 0; i < n; ++i)    {        dp[i] = 1;        scanf("%d", &m[i]);        for(int j = 0; j < i; j++)            if(m[j] < m[i]) dp[i] = max(dp[j]+1, dp[i]);        maxxx = max(maxxx, dp[i]);    }    printf("%d",maxxx);}int main(){    scanf("%d",&n)    lis();    return 0;}

2.合并石子
n 堆石子排成一圈，第i 堆石子有ai 个
每次可以合并相邻的两堆石子成一堆，代价为两堆石子的个数之和
求将n 堆石子合并成一堆的最小代价(n<=300)

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#define LL long long#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;const int maxn = 305;int f[maxn][maxn];int a[maxn],s[maxn];int n;int main(){    scanf("%d",&n);    for(int i = 1; i <= n; i++)        scanf("%d",&a[i]);    for(int i = 1; i <= n; i++)        s[i] = s[i-1] + a[i];    memset(f, 0x3f, sizeof(f));    for(int i = 1; i <= n; i++)        f[i][i] = 0;    for(int i = n; i >= 1; i--)        for(int j = i + 1; j <= n; j++)            for(int k = i; k <= j - 1; k++)                f[i][j] = min(f[i][j], f[i][k]+f[k+1][j]+s[j]-s[i-1]);    printf("%d\n",f[1][n]);    return 0;}

3.lcs 最长公共子串
已知字符串A、B,求前缀两两之间的最长公共子串。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#define LL long long#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std; const int maxn = 3005;char a[3][maxn];int dp[maxn][maxn];int len1,len2;int main(){    for(int i = 1; i <= 2; i++)        scanf("%s", a[i] + 1);    len1 = strlen(a[1] + 1);    len2 = strlen(a[2] + 1);    for(int i = 1; i <= len1; ++i) {        for(int j = 1; j <= len2; ++j) {            dp[i][j] = max(dp[i][j-1], dp[i-1][j]);            if(a[1][i] == a[2][j])            dp[i][j] = max(dp[i][j], dp[i-1][j-1] + 1);        }    }    printf("%d",dp[len1][len2]);    return 0;}

4.背包问题
……好多好多背包
就写个01、完全
①01背包
n 个物品，第i 个物品有体积vi，价值wi。
现在有容量为V 的背包，求可以容纳的物品的最大价值和。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#define LL long long#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std; const int maxm = 201, maxn = 31;int m,n;int w[maxn],c[maxn];int dp[maxm],f[maxn][maxm];void DP1(){    for(int i = 1; i <= n; i++)        for(int v = m; v > 0; v--)            if(w[i] <= v) f[i][v] = max(f[i-1][v], f[i-1][v-w[i]]+c[i]);            else f[i][v] = f[i-1][v];    printf("%d\n",f[n][m]);}void DP2(){    for(int i = 1; i <= n; i++)        for(int v = m; v >= w[i]; v--)            if(dp[v-w[i]] + c[i] > dp[v])                dp[v] = dp[v-w[i]] + c[i];    printf("%d\n",dp[m]);}int main(){    scanf("%d%d",&n,&m);    for(int i = 1; i <= n; i++)        scanf("%d%d",c[i],w[i]);    DP1();  //一维数组     DP2();  //二维数组     return 0;}

②完全背包

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#define LL long long#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std; const int maxm = 201, maxn = 31;int m,n;int w[maxn],c[maxn];int dp[maxm],f[maxn][maxm];void DP1(){    for(int i = 1; i <= n; i++)        for(int v = 1; v <= m; v++)            if(v < w[i]) f[i][v] = f[i-1][v];            else f[i][v] = max(f[i-1][v],f[i][v-w[i]]+c[i]);}void DP2(){    for(int i = 1; i <= n; i++)        for(int v = w[i]; v <= m; v++)            if(dp[v-w[i]]+c[i] > dp[v])                dp[v] = dp[v-w[i]]+c[i];    printf("%d\n",dp[m]);}int main(){    scanf("%d%d",&n,&m);    for(int i = 1; i <= n; i++)        scanf("%d%d",c[i],w[i]);    DP1();  //一维数组     DP2();  //二维数组     return 0;}