$ACM$ 课第四次作业-动态规划

\(ACM\) 课第四次作业-动态规划

---

\(A.\ Pearls\)

题意

给定 \(n\) 个珍珠，每个珍珠具有属性 \(a_{i},\ p_{i}\)，分别代表需求量和价格，珍珠按照品质由低到高升序列出

允许用高品质的珍珠代替低品质的珍珠

每次购买需要加上 \(10\) 个当前品质珍珠的价格

求买下所有需求的珍珠的最少金额

输入格式

多组数据，\(n\leq 100,\ a_{i},\ p_{i}\leq 1000\)

输出格式

输出最小金额

题解

观察发现，如果用高品质的珍珠代替低品质的珍珠，他们必须是连续的

考虑用第 \(j\) 个珍珠的价格组合购买 \(i,\ i + 1,\ ...\,\ j\) 的珍珠，单独购买第 \(k,\ i < k < j\) 个珍珠，假设这样的购买方式最优

完全可以用第 \(k\) 个珍珠的价格购买 \(i,\ i + 1,\ ...\,\ k\) 的珍珠，这样价格更少，所以组合购买必须连续

定义 \(dp[i]\) 是购买到第 \(i\) 个珍珠所花费的最少金额

则

\[dp[i] = max\left\{dp[j - 1],\ (sum[i] - sum[j - 1] + 10)\cdot p[i],\ 1\leq j\leq i\right\} \]

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 5e2 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;

inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}

int t, n, a[107], p[107], sum[107], dp[107];
int main()
{
    t = read();
    while(t--) {
        memset(dp, inf, sizeof(dp));
        dp[0] = 0;
        n = read();
        for(int i = 1; i <= n; ++i)
            a[i] = read(), p[i] = read();
        for(int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + a[i];
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= i; ++j) {
                dp[i] = min(dp[i], dp[j - 1] + (sum[i] - sum[j - 1] + 10) * p[i]);
            }
        }
        printf("%d\n", dp[n]);
    }
    return 0;
}
/*
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 -3 -2 5 0
1
10
3
-1 -5 -2
3
-1 0 -2
0
*/

---

\(B.\) 最大连续子序列

题意

给定长为 \(n\) 的序列，最大连续子序列，记录首尾

输入格式

多组数据，\(n\leq 10000\)

输出格式

输出和的最大值，以及对应的首尾

若有多个答案，按照字典序最小的输出

若和为负数，则输出 \(0,\ 1,\ n\)

题解

定义 \(dp[i]\) 为以 \(a[i]\) 结尾的子序列和的最大值

\[dp[i] = max\left\{dp[i - 1] + a[i],\ a[i]\right\} \]

记录最大值 \(ans\) 后，再扫一遍数组求首尾

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 5e2 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;

inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}

int t, n, a[10007], dp[10007];
int main()
{
    while(scanf("%d", &n) && n) {
        for(int i = 1; i <= n; ++i) a[i] = read();
        int ans = -inf, head = 0, tail = 0;
        for(int i = 1; i <= n; ++i) {
            dp[i] = max(dp[i - 1] + a[i], a[i]);
            ans = max(ans, dp[i]);
        }
        if(ans < 0) ans = 0, head = 1, tail = n;
        else {
            for(int i = 1; i <= n; ++i) {
                if(dp[i] == ans) {
                    int temp = ans;
                    head = tail = i;
                    for(int j = i; temp; temp -= a[j], j--)
                        head = j;
                    break;
                }
            }
        }
        printf("%d %d %d\n", ans, a[head], a[tail]);
    }
    return 0;
}
/*
6
-2 11 -4 13 -5 -2
10
-10 1 2 3 4 -5 -23 3 7 -21
6
5 -8 -3 -2 5 0
1
10
3
-1 -5 -2
3
-1 0 -2
0
*/

---

\(C.\ To\ The\ Max\)

题意

给定 \(N\times N\) 的矩阵，求最大子矩阵和

输入格式

多组数据，\(N\leq 100\)

输出格式

输出最大子矩阵和

题解

维护二维前缀和

枚举子矩阵左上角和右下角，用差分 \(O(1)\) 计算和

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 1e7 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;

inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}

int n, a[107][107], dp[107][107];
int main()
{
    while(cin >> n) {
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                a[i][j] = read();
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= n; ++j) {
                dp[i][j] = a[i][j] + dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1];
                //cout<<i<<' '<<j<<' '<<dp[i][j]<<endl;
            }
        }
        int ans = -inf;
        for(int s = 1; s <= n; ++s) {
            for(int t = 1; t <= n; ++t) {
                for(int i = 1; i <= s; ++i) {
                    for(int j = 1; j <= t; ++j) {
                        ans = max(ans, dp[s][t] - dp[i - 1][t] - dp[s][j - 1] + dp[i - 1][j - 1]);
                        //if(dp[s][t] - dp[i - 1][t] - dp[s][j - 1] + dp[i - 1][j - 1] == 23) cout<<i<<' '<<j<<' '<<s<<' '<<t<<endl;
                    }
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}
/*
4
-1 -1 -1 -1
-1 -1 -1 -1
-1 -1 -1 -1
-1 -1 -1 -1
*/

---

\(D.\ Piggy\ Bank\)

题意

\(n\) 个物品，每个物品有 \(p_{i},\ w_{i}\) 两个属性，分别代表金额与容量，可以取多次

给定容量为 \(W = F - E\) 的背包，求装满背包所需要的金额的最小值

输入格式

多组数据，\(n\leq 500,\ E,\ F,\ W\leq 10000,\ P\leq 50000\)

输出格式

若能恰好装满，输出最小金额

若不能恰好装满，输出“不可能”

题解

完全背包

\(code\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " \n"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 5e2 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;

inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}

int t, e, f, n, v[N], w[N], dp[10007];
int main()
{
    t = read();
    while(t--) {
        memset(dp, inf, sizeof(dp));
        dp[0] = 0;
        e = read(), f = read();
        n = read();
        for(int i = 1; i <= n; ++i) v[i] = read(), w[i] = read();
        for(int i = 1; i <= n; ++i)
            for(int j = w[i]; j <= (f - e); ++j)
                dp[j] = min(dp[j], dp[j - w[i]] + v[i]);
        //cout<<inf<<endl;
        if(dp[(f - e)] != inf) printf("The minimum amount of money in the piggy-bank is %d.\n", dp[(f - e)]);
        else puts("This is impossible.");
    }
    return 0;
}
/*
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4
*/