UVa 348

来源：互联网发布：九州通i9软件编辑：程序博客网时间：2024/05/21 11:12

矩阵连乘问题, 需要注意的是 () 的输出问题

现放出用递归写的一个RE代码, 递归这里会暴栈:

#include <stdio.h>#include <iostream>#include <string.h>using namespace std;#define MIN(A,B) (A<B?A:B)int path[15][15];int d[15][15];int a[15][2];int dp(int s, int e) {int i, j;int min, t, tp;min = 9999999;if(s >= e) return d[s][e] = 0;if(s+1 == e) return d[s][e] = a[s][0]*a[s][1]*a[e][1];if(d[s][e]) return d[s][e];for(i = s; i < e; i++) {tp = dp(s, i) + dp(i+1, e) + a[s][0]*a[s][1]*a[e][1];if(tp < min) {min = tp;path[s][e] = i;}}d[s][e] = tp;return d[s][e];}void print_path(int s, int e) {if(s == e) {printf("A%d", s);return;}if(s+1 == e) {printf("(A%d x A%d)", s, e);return;}printf("(");print_path(s, path[s][e]);printf("x");print_path(path[s][e]+1, e);printf(")");}int main() {int n;int cse = 1;while(scanf("%d", &n) != EOF) {int i, j;memset(d, 0, sizeof(d));memset(path, 0, sizeof(path));for(i = 1; i <= n; i++) {scanf("%d%d", &a[i][0], &a[i][1]);}dp(1, n);printf("Case %d: ", cse++);print_path(1, n);printf("\n");}return 0;}

非递归代码:

#include<stdio.h>#include<string.h>#define MAXD 20#define INF 1000000000int N, left[MAXD], right[MAXD], f[MAXD][MAXD], p[MAXD][MAXD];int init(){    int i;    scanf("%d", &N);    if(!N)        return 0;    for(i = 0; i < N; i ++)        scanf("%d%d", &left[i], &right[i]);}void printpath(int a, int b){    if(a == b)    {        printf("A%d", a + 1);        return ;    }    printf("(");    printpath(a, p[a][b]);    printf(" x ");    printpath(p[a][b] + 1, b);    printf(")");}void solve(){    int i, j, k, temp;    for(i = 0; i < N; i ++)        for(j = i; j < N; j ++)        {            if(i == j)                f[i][j] = 0;            else                f[i][j] = INF;        }    for(k = 1; k < N; k ++)        for(i = 0; i + k < N; i ++)            for(j = i; j < i + k; j ++)            {                temp = f[i][j] + f[j + 1][i + k] + left[i] * right[j] * right[i + k];                if(temp < f[i][i + k])                {                    f[i][i + k] = temp;                    p[i][i + k] = j;                }            }    printpath(0, N - 1);    printf("\n");}int main(){    int t = 0;    while(init())    {        printf("Case %d: ", ++ t);        solve();    }    return 0;}