【UVA】348-Optimal Array Multiplication Sequence（矩阵链乘）

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最基本的矩阵链乘。

一开始打错一个字母WA了N次，郁闷，打印路径的时候找到一条最佳路径不要忘记return。

递推式子为：

dp[i][j] = min(dp[i][k] + dp[k + 1][j] + mat[i].x * mat[k].y * mat[j].y);

348Optimal Array Multiplication SequenceAcceptedC++0.0662014-08-06 10:29:02

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<set>#include<ctime>#include<cmath>#include<string>#include<iomanip>#include<climits>#include<cctype>#include<deque>#include<list>#include<sstream>#include<vector>#include<cstdlib>using namespace std;#define _PI acos(-1.0)#define INF (1 << 30)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================矩阵链乘问题，d[i][j]代表i ~ j 编号内矩形相乘最小次数======================================*/#define MAXD 10 + 10int n,Case = 1;int dp[MAXD][MAXD];struct Mat{    int x;    int y;}mat[MAXD];void init(){    for(int i = 1 ; i <= n ; i++){        scanf("%d%d",&mat[i].x,&mat[i].y);    }    for(int i = 1 ; i <= n ; i++)        for(int j = i; j <= n ; j++)            if(i == j)                dp[i][j] = 0;            else                dp[i][j] = INF;    return;}void print(int i,int j){    if(i == j){        printf("A%d",i);        return ;    }    else    for(int k = i ; k < j ; k++)        if(dp[i][j] == dp[i][k] + dp[k + 1][j] + mat[i].x * mat[j].y * mat[k].y){            printf("(");print(i,k);printf(" x ");print(k + 1, j);printf(")");            return ;        }    return ;}void DP(){    init();    for(int j = 1 ; j <= n ; j++){        for(int i = 1 ; i + j <= n ; i++){             for(int k = i ; k < i + j ; k++)                dp[i][i + j] = min(dp[i][i + j],dp[i][k] + dp[k + 1][i + j] + mat[i].x * mat[i + j].y * mat[k].y);        }    }    printf("Case %d: ",Case++);    print(1,n);    printf("\n");    return ;}int main(){    while(scanf("%d",&n) && n){        DP();    }    return 0;}

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