UVA 442 Matrix Chain Multiplication
2015-07-28 08:43
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题目
矩阵链乘分析
给出n个矩阵的维度和一些矩阵链乘表达式,输出乘法的次数。如果乘法无法进行,输出error。
设A是m * n的矩阵,B是n * p的矩阵,那么AB是m * p的矩阵,乘法次数为m * n * p。如果A的列数不等于B的行数,则乘法无法进行。
思路
设置一个栈,遇到字母入栈,遇到右括号时将栈顶两个字母出栈并计算,再将结果入栈。代码
#include <cstdio> #define MAXN 200 int solve(char* buf, int* r, int* c) { int res = 0, top = 0; int mr[MAXN], mc[MAXN]; if (buf[1] == '\n') return 0; for (int i = 0; buf[i] != '\n'; i++) { if (buf[i] == ')') { /* pop */ if (mc[top-1] != mr[top]) return -1; /* error */ res += mr[top-1]*mc[top]*mr[top]; mc[top-1] = mc[top]; top--; } if (buf[i] >= 'A' && buf[i] <= 'Z') { /* push */ top++; mr[top] = r[buf[i]-'A']; mc[top] = c[buf[i]-'A']; } } return res; } int main() { int t, row[27], col[27], res; char ch, buf[MAXN]; scanf("%d\n", &t); while (t--) { scanf("%c", &ch); scanf(" %d %d\n", &row[ch-'A'], &col[ch-'A']); } while (fgets(buf, MAXN, stdin)) { /*printf("catch: %s", buf);*/ res = solve(buf, row, col); if (res >= 0) printf("%d\n", res); else printf("error\n"); } return 0; }
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