uva348 - Optimal Array Multiplication Sequence
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题意:
给你一些固定顺序的矩阵(行,列),求它们相乘所得到的最小乘积次数
Sample Input
3
1 5
5 20
20 1
3
5 10
10 20
20 35
6
30 35
35 15
15 5
5 10
10 20
20 25
0
Sample Output
Case 1: (A1 x (A2 x A3))
Case 2: ((A1 x A2) x A3)
Case 3: ((A1 x (A2 x A3)) x ((A4 x A5) x A6))
思路:
T_t,感觉dp题要不是做过同款题的话,我基本没什么思路,都是看了别人的题解,才得到启发。这题的话,可用区间dp来做。知道了后往这边尝试解答,还是退不出状态方程 。感觉大部分做不出来的题,都是没理解清楚这个过程,理解对过程的好过程再推公式,就是得心应手。dp题的思路实在是太重要了!
然后这题的话,主要就是区间dp,dp[i][j]的话就用来存下当前区间中矩阵的最小乘积次数,若该区间细分为(i,k),(k+1,j)后所得的乘积次数更小的话就更新,并用vis[i][j]=k来标记,即可解决输出问题。
代码如下:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 15;int dp[N][N], vis[N][N];int n;struct ju{ int x, y;}a[N];void print(int i, int j) { if (i == j) { printf("A%d", i + 1); return; } printf("("); print(i, vis[i][j]); printf(" x "); print(vis[i][j] + 1, j); printf(")");}int main() { int cas = 0; while (~scanf("%d", &n) && n) { for (int i = 0; i < n; i++) { scanf("%d%d", &a[i].x, &a[i].y); for (int j = 0; j < n; j++) { if (i == j) dp[i][j] = 0; else dp[i][j] = 0x3f3f3f3f; vis[i][j] = 0; } } for (int l = 1; l < n; l++) {//长 for (int i = 0; i < n; i++) { int j = i + l;//i~j //if (j > n) //break; //printf("%d %d:", i, j); for (int k = i; k < j; k++) {//i~k , k~j if (dp[i][j] > dp[i][k] + dp[k + 1][j] + a[i].x*a[k+1].x*a[j].y) { dp[i][j] = dp[i][k] + dp[k + 1][j] + a[i].x*a[k+1].x*a[j].y; vis[i][j] = k; //printf("%d",dp[i][j]); } } //printf("\n"); } } printf("Case %d: ", ++cas); print(0, n-1); printf("\n"); } return 0;}
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