爬格子呀7-2，例题7-6，7-11

上周写的，今天也写了一个，不过一直没调出来，改天发上来。
代码如下：
带宽：

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
set<int>neigh[26];
char tos[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
vector<int>tar;
vector<int>store;
int k = 8;

//操他妈迭代器用的飞起
int search() {
    int len, dis=0 ;
    vector<int>::iterator it;
    for (vector<int>::iterator i = tar.begin(); i != tar.end(); i++) {
        for (set<int>::iterator j = neigh[*i].begin(); j !=neigh[*i].end(); j++) {
            if (neigh[*i].size() >= k)
                return 0;
            it = find(tar.begin(), tar.end(), *j);
            len = abs(it - i);
            if (len >= k)
                return 0;
            if (len > dis)
                dis = len;
        }
    }
    return dis;
}

int main()
{
    char ch = getchar();
    int a = ch - 'A';
    set<int>mid;
    //记录谁的相邻节点有谁
    while ((ch = getchar()) != '#') {
        if (ch == ':')
            continue;
        else if (ch == ';') {
            ch = getchar();
            a = ch - 'A';
            mid.insert(ch - 'A');
        }
        else {
            neigh[a].insert(ch - 'A');
            neigh[ch - 'A'].insert(a);
            mid.insert(ch-'A');
        }
    }
    copy(mid.begin(), mid.end(), back_inserter(tar));
    int kase;
    do {
        kase = search();
        if (kase && kase < k) {
            k = kase;
            store.assign(tar.begin(), tar.end());
        }
    } while (next_permutation(tar.begin(), tar.end()));
    for (auto i : store) {
        cout << tos[i] << " ";
    }
    cout << "-> " << k;
    return 0;
}

1. 宝箱：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int s1, v1, s2, v2, n, kase, j;
long long assest;

int main() {
    cin >> kase;
    while (j++<kase) {
        scanf_s("%d%d%d%d%d", &n, &s1,&v1,&s2, &v2);
        if (n / s1 < 65536 || n / s2 < 65536) {
            assest = 0;
            if (n / s1 < n / s2) 
                for (long long i = 0; i <= n / s1; i++) 
                    assest = max((n - s1*i) / s2*v2 + i*v1, assest);
            else
                for (long long i = 0; i <= n / s2; i++) {
                    assest = max((n - s2*i) / s1*v1 + i*v2, assest);
                }
        }
        else {
            assest = 0;
            if (s1*v2 < s2*v1) { swap(s1, s2); swap(v1, v2); }
            for (long long i = 0; i <= s2 - 1; i++) {
                assest = max((n - i*s1) / s2*v2 + i*v1, assest);
            }
        }           
        printf("Case #%d: %lld
", j, assest);
    }
    return 0;
}

1. 例题7-2

#include<cstdio>
#include<vector>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
const int bd = 105;
const char tos[] = "news";
int g[250][250], v[21];
int dir[4][2] = { {1,0},{0,1},{0,-1},{-1,0} };
vector<int>path;
pair<int, int>_begin = { 0,0 };
int n, k, path_num;

void find_path() {
    for (int i = 0; i < path.size(); i++) {
        cout << tos[i] << " ";
    }
    cout << endl;
}

bool legal(int x, int y) {
    return abs(x) < bd && abs(y) < bd;
}

void dfs(pair<int, int>u) {
    if (path.size() == n) {
        if (u == _begin) {
            path_num++;
            find_path();
            return;
        }
    }
    int step = path.size();
    for (int i = 0; i < 4; i++) {
        if (step && (path[step - 1] + 1) % 4 / 2 == (i + 1) % 4 / 2)
            continue;//判断该往哪里转弯
        pair<int, int>p = u;
        bool flag = true;
        for (int j = 0; j < step; j++) {
            p.first += dir[i][0];
            p.second += dir[i][1];
            if (legal(p.first, p.second) || g[p.first + bd][p.second + bd] == -1) {
                flag = false;
                break;
            }
        }
        if (flag&&g[p.first + bd][p.second + bd] != 1) {
            path.push_back(i);
            g[p.first + bd][p.second + bd] = 1;
            dfs(p);
            //当程序运行到这一步的时候，说明递归调用的完毕
            //他运行出结果了，最好，接下来的操作也不会影响他什么
            //但是他如果没有运行出结果，接下来的操作就是非常必要的了
            g[p.first + bd][p.second + bd] = 0;//取消之前的标记
            path.pop_back();//等价于path。resize（），都是剔除最后之前插入的元素
        }
    }
}

int main() {
    cin >> n >> k;
    int a, b;
    memset(g, 0, sizeof(g));
    for (int i = 0; i < k; i++) {
        scanf_s("%d%d", &a, &b);
        g[a + bd][b + bd] = -1;
    }
    int step = 1;
    pair<int, int>u = { 0,0 };
    dfs(u);
    cout << path_num;
    return 0;
}