10149. 「一本通 5.1 例 3」凸多边形的划分

题意

给定一个具有 $N$ 个顶点的凸多边形,将顶点从 $1$ 至 $N$ 标号,每个顶点的权值都是一个正整数。将这个凸多边形划分成 $N-2$ 个互不相交的三角形,试求这些三角形顶点的权值乘积和至少为多少。

思路

首先随便搞一个多边形:

然后给它顺时针每个顶点表上序号:

然后枚举$i,j$,要求:$i+1<j$,然后给$i,j$连一条线,分割出来另一个多边形:多边形23456

然后在$i,j$范围内枚举$k$,使得多边形23456又可以分割。

分割成如下图:

设$f[i][j]$表示把$i,j$的多边形切割成三角形后的权值乘积之和的最小值。 可得:
$$f[i][j]=min{f[i][k]+f[k][j]+a[i]_a[j]_a[k]}(0<i<j<k\leq n)$$
初始化:
$$f[i][j]=inf(0<i\leq n,0<j\leq n)$$ $$f[i][i+1]=0(0<i<n)$$ 时间复杂度:$O(n^3)$ 输出结果:$f[1][n]$ 当然,这道题范围特别大:对于 $100\%$ 的数据,有 $N\le 50$,每个点权值小于 $10^9$。三个数相乘最高可达$10^{27}$,所以需要使用高精度。这里使用了C++大数类,转自代号4101

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
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000;
struct bign{
int d[maxn], len;

void clean() { while(len > 1 && !d[len-1]) len--; }

bign() { memset(d, 0, sizeof(d)); len = 1; }
bign(int num) { *this = num; }
bign(char* num) { *this = num; }
bign operator = (const char* num){
memset(d, 0, sizeof(d)); len = strlen(num);
for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
clean();
return *this;
}
bign operator = (int num){
char s[20]; sprintf(s, "%d", num);
*this = s;
return *this;
}

bign operator + (const bign& b){
bign c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] += b.d[i];
if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;
}
while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;
c.len = max(len, b.len);
if (c.d[i] && c.len <= i) c.len = i+1;
return c;
}
bign operator - (const bign& b){
bign c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] -= b.d[i];
if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;
}
while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;
c.clean();
return c;
}
bign operator * (const bign& b)const{
int i, j; bign c; c.len = len + b.len;
for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)
c.d[i+j] += d[i] * b.d[j];
for(i = 0; i < c.len-1; i++)
c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
c.clean();
return c;
}
bign operator / (const bign& b){
int i, j;
bign c = *this, a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
c.d[i] = j;
a = a - b*j;
}
c.clean();
return c;
}
bign operator % (const bign& b){
int i, j;
bign a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
a = a - b*j;
}
return a;
}
bign operator += (const bign& b){
*this = *this + b;
return *this;
}

bool operator <(const bign& b) const{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
if(d[i] != b.d[i]) return d[i] < b.d[i];
return false;
}
bool operator >(const bign& b) const{return b < *this;}
bool operator<=(const bign& b) const{return !(b < *this);}
bool operator>=(const bign& b) const{return !(*this < b);}
bool operator!=(const bign& b) const{return b < *this *this < b;}
bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}

string str() const{
char s[maxn]={};
for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
return s;
}
};
istream& operator >> (istream& in, bign& x){
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream& out, const bign& x){

out << x.str();
return out;
}
#define ll bign
ll f[55][55],a[55];
int n;
int main(){
cin>>n;
for(int i=1;i<=n;i+=1) cin>>a[i];
memset(f,63,sizeof(f));
for(int i=1;i<=n;i++) f[i][i+1]=0;
for(int L=2;L<=n-1;L++){
for(int i=1;i<=n-L;i++){
int j=i+L;
for(int k=i+1;k<=j-1;k++){
f[i][j]=min(f[i][k]+f[k][j]+a[i]*a[j]*a[k],f[i][j]);
}
}
}
cout<<f[1][n];putchar('\n');
return 0;
}
//f[i][j]=min{f[i][k]+f[k][j]+a[i]*a[j]*a[k]}(0<i<k<j<=n)
//f[i][j]=inf
//f[i][i+1]=0;
//end:f[1][n]
//Time:O(n^3)