动态规划综合(五)

这篇博文继续对动态规划的应用做一些探讨
还继续探讨了动态规划有关的优化问题

杂乱的dp

HDU2929

$\textbf{algorithm}$
可以用火柴作为 dp 的阶段,用余数作为状态集合
$d(i, j)$ 表示用 $i$ 根火柴,除以 $m$ 余数为 $j$
所能够摆出的最大正整数

$\textbf{start: } d(0,0) = 0$
$\textbf{end: for } \forall i \in [0, n] , \max d(i, 0)$

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
const int maxn = 100 + 10;
const int maxm = 3000 + 10;
const int maxl = 60;

// 10
const int mp[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};

int n, m;

char d[maxn][maxm + 1][maxl], ans[maxl];

inline int scmp(const char* a, const char* b) {
if(strlen(a) > strlen(b)) return 1;
else if(strlen(a) < strlen(b)) return -1;
return strcmp(a, b);
}

void work(char *a, const char *b, int k) {
char s[maxl];
strcpy(s, b);

int len = strlen(s);
if(len == 1 && s[len-1] == '0') {
s[len-1] = '0' + k;
s[len] = 0;
}
else {
s[len] = '0' + k;
s[len+1] = 0;
}

if(scmp(a, s) < 0) strcpy(a, s);
}

void initdp() {
memset(d, 0, sizeof(d));
d[0][0][0] = '0';
}

void dp() {
_rep(i, 0, n) {
_for(j, 0, maxm) if(strlen(d[i][j]) > 0) {
_for(k, 0, 10) work(d[ i+mp[k] ][(j*10 + k) % m], d[i][j], k);
}
}
}

void init() {
memset(ans, 0, sizeof(ans));
}

int main() {
freopen("input.txt", "r", stdin);
int kase = 0;

while (scanf("%d%d", &n, &m) == 2 && n) {
init();
printf("Case %d: ", ++kase);

// then solve the problem
initdp();
dp();

for(int i = n; i > 0; i--) if(scmp(ans, d[i][0]) < 0) strcpy(ans, d[i][0]);

//debug(ans);

if(ans[0] == 0) printf("-1\n");
else puts(ans);
}
}

字符串集合动态规划

LA3136

LA3136

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144

template <class T>
inline bool chmin(T& a, T b) {
if(a < 0 || b < a) {
a = b;
return true;
}
return false;
}

// ============================== //
const int maxn = 16;
const int maxl = 200 + 10;
const int inf = 0x3f3f3f3f;
int n;

class A {
public:
string s, rev;

void _init() {
rev = s;
reverse(rev.begin(), rev.end());
}

bool operator< (const A& rhs) const {
return s.length() < rhs.s.length();
}

} a[maxn];

string str[maxn][2];

void prework() {
sort(a, a + n);

int n2 = 0;
_for(i, 0, n) {
bool need = true;
_for(j, i + 1, n) {
if(a[j].s.find(a[i].s) != string::npos || a[j].rev.find(a[i].s) != string::npos) {
need = false;
break;
}
}

if(need) {
str[n2][0] = a[i].s;
str[n2][1] = a[i].rev;
//debug(strlen(str[n2][0].c_str()));
n2++;
}
}
n = n2;
}

void dbg() {
_for(i, 0, n) debug(str[i][0]), debug(str[i][1]);
}

int overlap[maxn+1][maxn+1][2][2];

int getOverlap(const string& a, const string& b) {
// default a < b
_for(i, 1, a.length()) {
if(i + b.length() <= a.length()) continue;

bool ok = true;
for(int k = 0; i + k < a.length() && k < b.length(); k++) if(b[k] != a[i+k]) {
ok = false;
break;
}

if(ok) return a.length() - i;
}

return 0;
}

void prework2() {
_for(i, 0, n) _for(j, 0, n) _for(x, 0, 2) _for(y, 0, 2) {
overlap[i][j][x][y] = getOverlap(str[i][x], str[j][y]);
}
}

int d[(1<<maxn)+1][maxn][2];

void initdp() {
memset(d, -1, sizeof(d));
d[1][0][0] = str[0][0].length();
assert(str[0][0].length() == str[0][1].length());
}

void dp() {
const int full = (1<<n) - 1;
_rep(s, 1, full) _for(j, 0, n) _for(dir, 0, 2) {
if(d[s][j][dir] >= 0) {
// place k
_for(k, 0, n) if(!(s & (1<<k))) {
assert(k != j);
_for(dir2, 0, 2) chmin(d[ s|(1<<k) ][k][dir2], d[s][j][dir] + (int)str[k][dir2].length() - overlap[j][k][dir][dir2]);
}
}
}
}

void get_ans() {
int ans = -1;
int full = (1<<n) - 1;
_for(i, 0, n) _for(dir, 0, 2) {
if(d[full][i][dir] < 0) continue;
chmin(ans, d[full][i][dir] - overlap[i][0][dir][0]);
}
if(ans <= 1) ans = 2;
printf("%d\n", ans);
}

void init() {
//_for(i, 0, maxn) _for(dir, 0, 2) str[i][dir].clear();
}

int main() {
freopen("input.txt", "r", stdin);
while (cin >> n && n) {
init();

// get data
_for(i, 0, n) {
cin >> a[i].s;
a[i]._init();
}
// get data finished

prework();
//dbg();
//debug("");

// then dp
prework2();
initdp();
dp();
get_ans();
}
}

单调队列优化dp

以经典的Fence为例
POJ1821

fence01
fence02

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89

template <class T>
inline bool chmax(T& a, T b) {
if(a < b) {
a = b;
return true;
}
return false;
}

template <class T>
inline bool chmin(T& a, T b) {
if(a > b) {
a = b;
return true;
}
return false;
}

// ============================================================== //

const int maxn = 16000 + 10;
const int maxm = 100 + 10;

int n, m;

class A {
public:
int _s, _l, _p;
A() {}
A(int s, int l, int p) : _s(s), _l(l), _p(p) {}

bool operator< (const A& rhs) const {
return _s < rhs._s;
}
} a[maxm];

int f[maxm][maxn];

void initdp() {
sort(a + 1, a + 1 + m);
memset(f, 0, sizeof(f));
}

inline int g(int i, int k) {
return f[i-1][k] - a[i]._p*k;
}

void dp() {
int que[maxn];
memset(que, 0, sizeof(que));

_rep(i, 1, m) {
int l = 1, r = 0;
for(int k = max(0, a[i]._s-a[i]._l); k <= a[i]._s-1; k++) {
while (l <= r && g(i, que[r]) <= g(i, k)) r--;
que[++r] = k;
}

_rep(j, 1, n) {
f[i][j] = max(f[i - 1][j], f[i][j - 1]);
if(j >= a[i]._s) {
while (l <= r && que[l] < j - a[i]._l) l++;
if(l <= r) chmax(f[i][j], a[i]._p*j + g(i, que[l]));
}
}
}
}

void init() {
//
}

int main() {
freopen("input.txt", "r", stdin);
cin >> n >> m;
init();

// get data
_rep(i, 1, m) {
scanf("%d%d%d", &a[i]._l, &a[i]._p, &a[i]._s);
}

// then dp
initdp();
dp();
int ans = f[m][n];
printf("%d\n", ans);
}

滑动窗口和单调队列优化dp

POJ3017
POJ3017

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74

const int maxn = 1e5 + 10;
int n;
ll M;
ll a[maxn];

multiset<ll> st;
multiset<ll>::iterator it;

ll f[maxn];

void initdp() {
memset(f, 0, sizeof(f));
}


void dp() {
ll sum = 0;
int que[maxn];
memset(que, 0, sizeof(que));

int l = 1, r = 0;
int j = 0;

_rep(i, 1, n) {
sum += a[i];
while (sum > M) sum -= a[j++];
//debug(j);

while (l <= r && que[l] < j) {
int x = que[l++];
if(l <= r && (it = st.find( f[x]+a[que[l]] )) != st.end()) st.erase(it);
}

while (l <= r && a[que[r]] <= a[i]) {
int x = que[r--];
if(l <= r && (it = st.find( f[que[r]]+a[x] )) != st.end()) st.erase(it);
}

if(l <= r) st.insert( f[que[r]]+a[i] );
que[++r] = i;

f[i] = f[j - 1] + a[que[l]];
//debug(*st.begin());
if(st.begin() != st.end()) f[i] = min(f[i], *st.begin());
//debug(f[i]);
}
}

void init() {
st.clear();
}

int main() {
freopen("input.txt", "r", stdin);
cin >> n >> M;
init();

_rep(i, 1, n) scanf("%lld", &a[i]);
//debug(a[1]);

_rep(i, 1, n) {
if(a[i] > M) {
puts("-1");
return 0;
}
}

// then dp
initdp();
dp();
ll ans = f[n];
printf("%lld\n", ans);
}

单调性和决策集思想

01

dp杂题

HDU1514

一开始不好想,但很明显应该用 $dp(x_1, x_2, x_3, x_4)$ 作为记忆化搜素
其中 $dp$ 的参数表示堆的高度

$\text{限制条件有两个,第一是 } cnt = 5? $
$\text{第二是 } canTake(i) \text{ 表示能否从第 i 堆拿糖果} $
$\text{注意保证第 i 堆有糖果剩余,才可以} $

$v[…] \rightarrow v’[…]$ 表示此时篮子里的糖果状态,每种颜色的糖果有还是没有?

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76

// ============================================================== //

const int maxn = 40 + 10;
int a[maxn][maxn];
int n;

int top[maxn];
int dp[maxn][maxn][maxn][maxn];

bool cantake(int i) {
return top[i] > 0;
}

int vis[maxn];

void initdp() {
_rep(i, 0, 4) top[i] = n;
assert(top[0] >= 0);
memset(dp, -1, sizeof(dp));
memset(vis, 0, sizeof(vis));
}

int dfs(int cnt, int vis[]) {
int& ans = dp[top[0]][top[1]][top[2]][top[3]];
if(ans >= 0) return ans;

if(cnt == 5) {
ans = 0;
return ans;
}

ans = 0;
for(int i = 0; i < 4; i++) {
if(!cantake(i)) continue;

int color = a[i][top[i]];
top[i]--;

if(!vis[color]) {
vis[color] = 1;
chmax(ans, dfs(cnt+1, vis));
vis[color] = 0;
}
else {
vis[color] = 0;
chmax(ans, dfs(cnt-1, vis) + 1);
vis[color] = 1;
}

top[i]++;
}

return ans;
}

void init() {
//
}

int main() {
freopen("input.txt", "r", stdin);
while (cin >> n && n) {
init();
_for(j, 0, n) _for(i, 0, 4) {
int x;
scanf("%d", &x);
a[i][n-j] = x;
}

// then dp
initdp();
int ans = dfs(0, vis);
printf("%d\n", ans);
}
}

单调队列优化多重背包

package01
package02
package03
package04

HDU2191

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67

const int maxn = 20000 + 10;
const int inf = 0x3f3f3f3f;
int n, m;

int v[maxn], w[maxn], c[maxn];

int f[maxn];
int que[maxn];
void initdp() {
memset(f, -inf, sizeof(f));
memset(que, 0, sizeof(que));
f[0] = 0;
}

inline int calc(int i, int u, int k) {
return f[u + k*v[i]] - k*w[i];
}

void dp() {
_rep(i, 1, n) {
_rep(u, 0, v[i] - 1) {
int l = 1, r = 0;
int D = (m-u) / v[i];

for(int k = D-1; k >= max(0, D-c[i]); k--) {
while (l <= r && calc(i, u, que[r]) <= calc(i, u, k)) r--;
que[++r] = k;
}

for(int p = D; p >= 0; p--) {
while (l <= r && que[l] > p-1) l++;
if(l <= r) chmax(f[u + p*v[i]], calc(i, u, que[l]) + p*w[i]);

if(p - c[i] - 1 >= 0) {
while (l <= r && calc(i, u, que[r]) <= calc(i, u, p-c[i]-1)) r--;
que[++r] = p - c[i] - 1;
}
}
}
}

int ans = 0;
_rep(i, 1, m) ans = max(ans, f[i]);
printf("%d\n", ans);
}

void init() {
//
}

int main() {
freopen("input.txt", "r", stdin);
int kase;
cin >> kase;
while (kase--) {
init();
scanf("%d%d", &m, &n);
_rep(i, 1, n) {
scanf("%d%d%d", &v[i], &w[i], &c[i]);
}

// then dp
initdp();
dp();
}
}