爬格子呀7-2，例题7-6，7-11

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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;}

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\n", j, assest);    }    return 0;}

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;}

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