UVA - 442 Matrix Chain Multiplication 双端队列

题目大意：给出n个矩阵和表达式，问该表达式是否正确，如果计算正确，输出计算了多少次

解题思路：双端队列，遇到右括号时弹出后面的两个矩阵进行乘法，相乘时要注意顺序，是第二个出队列的乘上第一个出队列的。

#include<cstdio>
#include<algorithm>
#include<deque>
#include<map>
#include<cstring>
using namespace std;
#define maxn 50010

struct Matrix{
    int x, y;
    Matrix() {}
    Matrix(int xx, int yy) {
        x = xx;
        y = yy;
    }
}M[30];

int n;
char str[maxn];
map<char, Matrix> Map;

void solve() {
    deque<Matrix> dq;
    int len = strlen(str);
    bool flag = true;
    long long ans = 0;

    for(int i = 0; i < len; i++) {
        if(str[i] == ‘ ‘ || str[i] == ‘(‘)
            continue;
        else if(str[i] == ‘)‘) {
            Matrix t2 = dq.back();
            dq.pop_back();
            Matrix t1 = dq.back();
            dq.pop_back();

            if(t1.y != t2.x) {
                flag = false;
                break;
            }
            ans += t1.x * t1.y * t2.y;
            dq.push_back(Matrix(t1.x, t2.y));
        }
        else {
            dq.push_back(Map[str[i]]);
        }
    }

    while(!dq.empty()) {
        Matrix t1 = dq.front();
        dq.pop_front();
        if(dq.empty()) 
            break;
        Matrix t2 = dq.front();
        dq.pop_front();
        if(t1.y != t2.x) {
            flag = false;
            break;
        }

        ans += t1.x * t1.y * t2.y;
        dq.push_front(Matrix(t1.x, t2.y));
    }

    if(!flag)
        printf("error\n");
    else
        printf("%lld\n", ans);
}
int main() {
    int n, x, y;
    char t, in[100];
    gets(in);
    sscanf(in,"%d",&n);
    for(int i = 0; i < n; i++) {
        gets(in);
        sscanf(in,"%c%d%d", &t, &x, &y);
        Map[t] = Matrix(x,y);
    }

    while(gets(str)) { 
        solve();
    }
    return 0;
}