(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，(nleq 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] = maxleft{dp[j - 1], (sum[i] - sum[j - 1] + 10)cdot p[i], 1leq jleq i
ight}
]
(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], "
"[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
", 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，求最大连续子序列的和，并且记录首尾
输入格式
多组数据，(nleq 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] = maxleft{dp[i - 1] + a[i], a[i]
ight}
]
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], "
"[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
", 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 imes N) 的矩阵，求最大子矩阵和
输入格式
多组数据，(Nleq 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], "
"[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];
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]);
printf("%d
", 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，(nleq 500, E, F, Wleq 10000, Pleq 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], "
"[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.
", 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
*/