Hlg 【表达式求值+欧拉路径】.cpp

题意：

　　给出一些表达式..求出表达式的和..

　　然后判断这些表达式是否可以组成一串首尾相连的数字串..

　　P.S. 如果有多组答案则按字典序给出..

 

思路：

　　栈的应用求出表达式..

　　深搜求欧拉路径..

　　

Tips：

　　如果有可行解则在深搜过程到尾的时候就输出~

　　否则的话因为是栈的应用~可能返回的过程中会导致答案的更改..

 

Code：

#include <stdio.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <ctype.h>
#include <map>
using namespace std;

struct Node
{
    char arr[1024];
    int st;
    int en;
}node[1024];

const int MAXN = 40;
int com[250][250];

void ini()
{
    memset(com, 0xff, sizeof(com));
    com['+']['+'] = com['+']['-'] = com['+'][')'] = com['+']['#'] = 1;
    com['-']['-'] = com['-']['+'] = com['-'][')'] = com['-']['#'] = 1;
    com['*']['+'] = com['*']['-'] = com['*']['*'] = com['*']['/'] = com['*'][')'] = com['*']['#'] = 1;
    com['/']['+'] = com['/']['-'] = com['/']['*'] = com['/']['/'] = com['/'][')'] = com['/']['#'] = 1;
    com[')']['+'] = com[')']['-'] = com[')']['*'] = com[')']['/'] = com[')'][')'] = com[')']['#'] = 1;
    com['('][')'] = com['#']['#'] = 0;
}

int change(char *&arr)
{
    int sum = 0;
    while(isdigit(*arr)) {
        sum = sum * 10 + (*arr-'0');
        arr++;
    }
    return sum;
}

int cal(int a, char o, int b)
{
    if(o == '+') return a+b;
    else if(o == '-') return a-b;
    else if(o == '*') return a*b;
    else if(o == '/') return a/b;
}

bool isoptr(char c) {
    if(c == '(' || c == '+' || c == '-' || c == '*' || c == '/' || c == ')' || c == '#') return true;
    else return false;
}

int calculate(char *arr)
{
    char optr[MAXN];
    int opnd[MAXN];
    char c;
    int a, b;
    int top1 = 0, top2 = 0;
    optr[top1++] = '#';
    while(*arr != '#' || optr[top1-1] != '#') {
        if(!isoptr(*arr)) {
           opnd[top2++] = change(arr);
          // arr++;//!!!
        } else {
            c = optr[top1-1];
            switch(com[c][*arr]) {
                case -1:
                    optr[top1++] = *arr;
                    arr++;
                    break;
                case 0:
                    --top1;
                    arr++;
                    break;
                case 1:
                    c = optr[--top1];
                    b = opnd[--top2];
                    a = opnd[--top2];
                    opnd[top2++] = cal(a, c, b);
                    break;
            }
        }
    }
    return opnd[top2-1];
}

bool vis[1010];
int val[1010];
int stack[1010], top;
bool flag;
int n;

void dfs(int u, int tot) {
    if(flag)
        return;
    if(tot == n) {
        flag = true;
        for(int i = 0; i < n; ++i) {
            printf("%d%c", val[stack[i]], i == n-1?'\n':'.');
        }
        return;
    }
    for(int i = 0; i < n; ++i)
    if(node[u].en == node[i].st && !vis[i]) {
        stack[top++] = i;
        vis[i] = true;
        dfs(i, tot+1);
        --top;
        vis[i] = false;
    }

}
bool cmp(Node a, Node b)
{
    return strcmp(a.arr, b.arr)<=0;
}
int main()
{
    int i, j, k;
    int T;
    char arr[100];
    while(scanf("%d", &T) != EOF)
    for(k = 1; k <= T; ++k) {
        scanf("%d", &n);

        memset(vis, 0, sizeof(vis));
        top = 0;
        ini();
        flag = false;

        for(i = 0; i < n; ++i) {
            scanf("%s", arr);
            val[i] = calculate(arr);
        }

        printf("Case #%d:\n", k);
        for(i = 0; i < n; ++i) {
            sprintf(node[i].arr, "%d", val[i]);
            node[i].st = node[i].arr[0]-'0';
            node[i].en = val[i]%10;

        }

        sort(node, node+n, cmp);
        for(i = 0; i < n; ++i) {
            vis[i] = true;
            stack[top++] = i;
            dfs(i, 1);
            if(flag) break;
            vis[i] = false;
            --top;
        }

        if(!flag) puts("NO DIGIT SEQUENCE");
        puts("");

    }
    return 0;
}

 

题目链接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1523