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  • UVA 140 Bandwidth(带宽)(全排列)

    题意:给定图,求是带宽最小的结点排列。

    分析:结点数最多为8,全排列即可。顶点范围是A~Z。

    #pragma comment(linker, "/STACK:102400000, 102400000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<cmath>
    #include<iostream>
    #include<sstream>
    #include<iterator>
    #include<algorithm>
    #include<string>
    #include<vector>
    #include<set>
    #include<map>
    #include<stack>
    #include<deque>
    #include<queue>
    #include<list>
    #define Min(a, b) ((a < b) ? a : b)
    #define Max(a, b) ((a < b) ? b : a)
    typedef long long ll;
    typedef unsigned long long llu;
    const int INT_INF = 0x3f3f3f3f;
    const int INT_M_INF = 0x7f7f7f7f;
    const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
    const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
    const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
    const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
    const int MOD = 1e9 + 7;
    const double pi = acos(-1.0);
    const double eps = 1e-8;
    const int MAXN = 100 + 10;
    const int MAXT = 10000 + 10;
    using namespace std;
    char s[MAXN];
    set<int> node[30];
    vector<int> v;
    int t[10];
    void init(){
        for(int i = 0; i < 30; ++i){
            node[i].clear();
        }
        v.clear();
        memset(t, 0, sizeof t);
        int len = strlen(s);
        for(int i = 0; i < len; ++i){
            if(s[i] == ':'){
                int tmp = s[i - 1] - 'A';
                while(1){
                    ++i;
                    if(i == len) break;
                    if(s[i] == ';') break;
                    node[tmp].insert(s[i] - 'A');
                    node[s[i] - 'A'].insert(tmp);
                }
            }
        }
    }
    int main(){
        while(scanf("%s", s) == 1){
            if(s[0] == '#') return 0;
            init();
            for(int i = 0; i < 30; ++i){
                if(node[i].size()){
                    v.push_back(i);
                }
            }
            int len = v.size();
            int ans = INT_M_INF;
            do{
                int tmp = 0;
                for(int i = 0; i < len; ++i){
                    for(int j = 0; j < i; ++j){
                        if(node[v[i]].count(v[j])){
                            tmp = max(tmp, i - j);
                        }
                    }
                }
                if(tmp < ans){
                    ans = tmp;
                    for(int i = 0; i < len; ++i){
                        t[i] = v[i];
                    }
                }
            }while(next_permutation(v.begin(), v.end()));
            for(int i = 0; i < len; ++i){
                printf("%c ", 'A' + t[i]);
            }
            printf("-> %d\n", ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/tyty-Somnuspoppy/p/6288041.html
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