Created
August 16, 2013 09:19
-
-
Save soulmachine/6248459 to your computer and use it in GitHub Desktop.
wikioi 1154, 能量项链, http://www.wikioi.com/problem/1154
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
/* wikioi 1154, 能量项链, http://www.wikioi.com/problem/1154 */ | |
#include <stdio.h> | |
#include <memory.h> | |
#include <limits.h> | |
#define MAXN 101 | |
int N; /** 矩阵的个数. */ | |
int a[MAXN]; | |
int p[MAXN]; /** 矩阵Ai的维度是p[i-1]xp[i]. */ | |
int d[MAXN][MAXN]; /** 状态,d[i][j]表示子问题Ai~Aj 的最优解. */ | |
int s[MAXN][MAXN]; /** 子问题Ai~Aj 应该在s[i][j]处断开 */ | |
/** | |
* @brief 打印子问题Ai~Aj的解 | |
* @param[in] i Ai | |
* @param[in] j Aj | |
* @return 无 | |
*/ | |
void print(const int i, const int j) { | |
if (i == j) { | |
printf("A%d",i); | |
} else { /* i < j */ | |
printf("("); | |
print(i, s[i][j]); | |
printf(" x "); | |
print(s[i][j]+1, j); | |
printf(")"); | |
} | |
} | |
void dp() { | |
int i, j, k, l; /* l表示区间长度 */ | |
//for (i = 1;i <= N; ++i) d[i][i]=0; | |
memset(d, 0, sizeof(d)); | |
for (l = 2; l <= N; ++l) { | |
for (i = 1; i <= N - l + 1; ++i) { | |
j = i + l -1; | |
d[i][j] = INT_MIN; | |
for (k = i; k < j; ++k) { | |
/* 大于号改为小于号 */ | |
if (d[i][j] < d[i][k] + d[k+1][j] + p[i-1] * p[k] * p[j]) { | |
d[i][j] = d[i][k] + d[k+1][j] + p[i-1] * p[k] * p[j]; | |
s[i][j] = k; | |
} | |
} | |
} | |
} | |
} | |
int main() { | |
int i, j; | |
if (scanf("%d", &N) && N > 0) { | |
for (i = 0; i < N; ++i) scanf("%d", &a[i]); | |
int max = INT_MIN; | |
j = 0; | |
for (j = 0; j < N; ++j) { | |
memset(s, 0, sizeof(s)); | |
for (i = 0; i < N; ++i) { | |
p[i] = a[(i+j)%N]; | |
p[i+1] = a[(i+j+1)%N]; | |
} | |
dp(); | |
//print(1, N); | |
//printf("%d\n", d[1][N]); | |
max = max > d[1][N] ? max : d[1][N]; | |
} | |
printf("%d\n", max); | |
} | |
return 0; | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment