2016 计蒜之道 微软的员工福利 (简单、中等)(dp)

题意:

$给定一颗N个节点的树,根为1,每个点可以在2种物品价值中2选1$
$每个节点会减少一定的价值f_i,其中x_i为所有直接儿子选择物品的极差(最大值-最小值)$
$$f_i=\lceil{x_i\over 1000}\rceil\times 666\times i $$
$求所有员工最大满意度的和$

分析:

$对于N\le 15的,直接二进制枚举谁拿第一种就好了,然后就做完了$
$对于N\le 100的:$
$显然的状态f[u][c]:=以u为根的子树,且这个点颜色是c(c=0第一种)的最大满意度和$
$转移的时候就枚举最大值和最小值,他们的个数都是O(n)个的,所以转移O(n^2)$
$总时间复杂度为O(n^3)$
$细节比较多,需要仔细,具体看代码$
$对于N\le 10^5的不会做。。其实看出来这个转移冗余很多,需要更多的性质,题解好神。$

代码一:

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
//
// Created by TaoSama on 2016-06-11
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
typedef long long LL;
int n;
vector<int> G[N], T[N];
int r[N], p[N];
void dfs(int u, int f) {
for(int v : G[u]) {
if(v == f) continue;
T[u].push_back(v);
dfs(v, u);
}
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
clock_t _ = clock();
while(scanf("%d", &n) == 1) {
for(int i = 0; i < n; ++i) {
scanf("%d%d", r + i, p + i);
G[i].clear();
T[i].clear();
}
for(int i = 1; i < n; ++i) {
int x, y; scanf("%d%d", &x, &y);
--x; --y;
G[x].push_back(y);
G[y].push_back(x);
}
dfs(0, -1);
LL ans = -1e18;
for(int i = 0; i < 1 << n; ++i) {
int b[20] = {};
for(int j = 0; j < n; ++j)
if(i >> j & 1) b[j] = r[j];
else b[j] = p[j];
LL tmp = 0;
for(int j = 0; j < n; ++j) {
int maxv, minv;
maxv = minv = b[j];
for(int son : T[j]) {
maxv = max(maxv, b[son]);
minv = min(minv, b[son]);
}
LL diff = maxv - minv;
LL delta = 1LL * (diff + 999) / 1000 * 666 * (j + 1);
tmp += b[j] - delta;
}
ans = max(ans, tmp);
}
printf("%lld\n", ans);
}
#ifdef LOCAL
printf("\nTime cost: %.2fs\n", 1.0 * (clock() - _) / CLOCKS_PER_SEC);
#endif
return 0;
}

代码二:

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
//
// Created by TaoSama on 2016-06-12
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, val[N][2];
vector<int> G[N];
int f[N][2];
void dfs(int u, int fa) {
f[u][0] = f[u][1] = 0;
vector<int> sons;
for(int v : G[u]) {
if(v == fa) continue;
sons.push_back(v);
dfs(v, u);
}
for(int i = 0; i < 2; ++i) {
int x = val[u][i], cur = -INF;
vector<int> big, small;
big.push_back(u); small.push_back(u);
for(int v : sons) {
if(val[v][0] >= x || val[v][1] >= x)
big.push_back(v);
if(val[v][0] <= x || val[v][1] <= x)
small.push_back(v);
}
for(int a : big) {
for(int j = 0; j < 2; ++j) {
if(a == u && i != j) continue;
int maxv = val[a][j];
if(maxv < x) continue;
for(int b : small) {
for(int k = 0; k < 2; ++k) {
if(b == u && i != k) continue;
if(a == b && j != k) continue;
int minv = val[b][k];
if(minv > x) continue;
int tmp = f[a][j], ok = 1;
if(a != b) tmp += f[b][k]; //fuck
for(int v : sons) {
if(v == a || v == b) continue;
int choice = -INF;
for(int c = 0; c < 2; ++c) {
int y = val[v][c];
if(y < minv || y > maxv) continue;
choice = max(choice, f[v][c]);
}
if(choice == -INF) {ok = 0; break;}
tmp += choice;
}
if(!ok) continue;
int diff = maxv - minv;
int delta = (diff + 999) / 1000 * 666 * u;
cur = max(cur, tmp + x - delta);
}
}
}
}
f[u][i] = cur;
}
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
clock_t _ = clock();
while(scanf("%d", &n) == 1) {
for(int i = 1; i <= n; ++i)
for(int j = 0; j < 2; ++j)
scanf("%d", val[i] + j);
for(int i = 1; i <= n; ++i) G[i].clear();
for(int i = 1; i < n; ++i) {
int u, v; scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1, -1);
// for(int i = 1; i <= n; ++i) {
// for(int j = 0; j < 2; ++j)
// printf("f[%d][%d] = %d\n", i, j, f[i][j]);
// }
printf("%d\n", max(f[1][0], f[1][1]));
}
#ifdef LOCAL
printf("\nTime cost: %.2fs\n", 1.0 * (clock() - _) / CLOCKS_PER_SEC);
#endif
return 0;
}


1. 除非注明,本博文即为原创,转载请注明链接地址
2. 本博文只代表博主当时的观点或结论,请不要恶意攻击
3. 如果本文帮到了您,不妨点一下 下面分享到 按钮,让更多的人看到