UVA 348 - Optimal Array Multiplication Sequence
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这道题的输出路径是个问题,仔细体会。
代码:
#include<stdio.h>
#include<string.h>
#define MAXN 10 + 5
int A[MAXN][2], f[MAXN][MAXN],part1[MAXN][MAXN];
int n;
void printpath(int a, int b)
{
if(a == b)
{
printf("A%d", a + 1);
return ;
}
printf("(");
printpath(a, part1[a][b]);
printf(" x ");
printpath(part1[a][b] + 1, b);
printf(")");
}
void dp()
{
memset(f,0,sizeof(f));
//memset(part1,-1,sizeof(part1));
for(int i = 1; i < n; i ++)
for(int j = 0; j < n-i; j ++)
{
for(int k = j; k < i+j; k ++)
{
if(k == j)
{
f[j][i+j] = f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[i+j][1];
part1[j][i+j] = k;
}
if(f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[j+i][1] < f[j][i+j])
{
f[j][i+j] = f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[j+i][1];
part1[j][i+j] = k;
}
}
}
}
void input()
{
int num = 0;
while(scanf("%d",&n) == 1)
{
if(n == 0) break;
for(int i = 0; i < n; i ++)
scanf("%d%d",&A[i][0],&A[i][1]);
dp();
printf("Case %d: ",++ num);
printpath(0,n-1);
puts("");
}
}
int main()
{
input();
return 0;
}
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