Bandwidth UVA

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题目描述：给定一个字符串说明各字母之间相连的情况，每个字母的距离为到与它相连的字母中的最大距离，对这些字母求一个序列，使得最大距离最小。

思路：全排列么枚举，求出每种情况的最大距离取最小值即可。比赛时脑残读错题意了，以为排列顺序只能按边走结果dfs了半天都推不出样例中的3。改了一个小时反应过来，又删了用全排列写，结果全排列忘记排序并且全排列格式错了，一直不出结果，无奈又改回dfs模拟全排列，匆忙写完看都没看离比赛结束还有一分半的时候交上1Y过了。其实是有点小郁闷的，要不是读错题这题也不会写了将近两个小时。

代码：下面附上全排列和dfs两个版本的代码，vj上显示全排列的时间更短。

全排列版本：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int maxn = 30;const int maxt = 1e6 + 10;const int mod = 10000007;const int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;int n, m, k;map<char, int> mp;map<int, char> mp2;vector<int> g[maxn];int d[maxn][maxn];int num[maxn];bool vis[maxn];int total_num;string str;int order[maxn];int ans_order[maxn];int ans;int total_len;void init(){    int cnt = 0;    for(char ch = 'A'; ch <= 'Z'; ++ch){//每个大写字母对应一个ID值        mp[ch] = ++cnt;    }    char ch = 'A';    for(int i = 1; i <= 26; ++i){        mp2[i] = ch;        ++ch;    }}int main(){    init();    while(cin >> str){        if(str == "#") break;        for(int i = 0; i < maxn; ++i) g[i].clear();        char cu,  cv;        int u, v;        int len = str.size();        int cur;        int cnt = 1;        string tmp;        memset(d, -1, sizeof d);        for(int i = 0; i < len; ++i){            cu = str[i];            cur = str.find(';', i);            if(cur == -1) cur = len;            i += 2;            for(; i < cur; ++i){                cv = str[i];                g[mp[cu]].push_back(mp[cv]);//保存边                g[mp[cv]].push_back(mp[cu]);                d[mp[cu]][mp[cv]] = d[mp[cv]][mp[cu]] = 1;//标记两边相连            }        }        total_num = mp.size();        total_len = 0;        for(int i = 1; i <= total_num; ++i){            if(!g[i].empty())sort(g[i].begin(), g[i].end()), ++total_len;//对于每个点对与其相连的点排序        }        ans = inf;        cnt = 0;        for(int i = 1; i <= total_num; ++i){            if(!g[i].empty())            order[++cnt] = i;        }        do{            int tmp = 0, cur;            for(int i = 1; i <= total_len; ++i){                for(int j = total_len; j >= i; --j)                    if(d[order[i]][order[j]] != -1){                        tmp = tmp < j - i ? j - i : tmp;//找该序列中的最大距离                        if(tmp >= ans) break;                    }            }            if(tmp >= ans) continue;//保存最小值            else{                ans = tmp;                for(int i = 1; i <= total_len; ++i){                    ans_order[i] = order[i];                }            }        }while(next_permutation(order + 1, order + 1 + total_len));//全排列枚举所有情况        for(int i = 1; i <= total_len; ++i){            printf("%c ", mp2[ans_order[i]]);        }        printf("-> %d\n", ans);    }    return 0;}

DFS版本：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int maxn = 30;const int maxt = 1e6 + 10;const int mod = 10000007;const int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;int n, m, k;map<char, int> mp;map<int, char> mp2;vector<int> g[maxn];int d[maxn][maxn];int num[maxn];bool vis[maxn];int total_num;string str;int order[maxn];int ans_order[maxn];int ans;int total_len;void init(){    int cnt = 0;    for(char ch = 'A'; ch <= 'Z'; ++ch){        mp[ch] = ++cnt;    }    char ch = 'A';    for(int i = 1; i <= 26; ++i){        mp2[i] = ch;        ++ch;    }}void solve(int u, int cnt){    if(cnt > total_len) return;    if(cnt == total_len){        int tmp = 0, cur;        for(int i = 0; i < cnt; ++i){//找最大距离                for(int j = cnt - 1; j >= 0; --j)                if(d[order[i]][order[j]] != -1){                    tmp = tmp < j - i ? j - i : tmp;                }        }        if(ans > tmp){//记录最大距离中的最小值及对应排列顺序            ans = tmp;            for(int i = 0; i < cnt; ++i){                ans_order[i] = order[i];            }        }    }    int v, l = g[u].size();    for(int i = 1; i <= total_num; ++i){        if(g[i].empty() || vis[i]) continue;        vis[i] = true;        order[cnt] = i;        solve(i, cnt + 1);        vis[i] = false;    }}int main(){    init();    while(cin >> str){        if(str == "#") break;        for(int i = 0; i < maxn; ++i) g[i].clear();        char cu,  cv;        int u, v;        int len = str.size();        int cur;        int cnt = 1;        string tmp;        memset(d, -1, sizeof d);        for(int i = 0; i < len; ++i){            cu = str[i];            cur = str.find(';', i);            if(cur == -1) cur = len;            i += 2;            for(; i < cur; ++i){                cv = str[i];                g[mp[cu]].push_back(mp[cv]);                g[mp[cv]].push_back(mp[cu]);                d[mp[cu]][mp[cv]] = d[mp[cv]][mp[cu]] = 1;            }        }        total_num = mp.size();        total_len = 0;        for(int i = 1; i <= total_num; ++i){            if(!g[i].empty())sort(g[i].begin(), g[i].end()), ++total_len;        }        ans = inf;        for(int i = 1; i < total_num; ++i){//以每个点（即每个字母）为起点枚举排列顺序。            if(g[i].empty()) continue;            memset(vis, 0, sizeof vis);            memset(order, 0, sizeof order);            vis[i] = true;            order[0] = i;            solve(i, 1);        }        for(int i = 0; i < total_len; ++i){            printf("%c ", mp2[ans_order[i]]);        }        printf("-> %d\n", ans);    }    return 0;}

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