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  • POJ 1094 Sorting It All Out

    题目链接:https://vjudge.net/problem/POJ-1094

    题目大意

       给定 N 个变量的 M 个大小关系,依次处理,判断这 N 个变量的大小关系能否确定?如能确定,在处理第几个大小关系时检测成功?如不能确定,是否有矛盾?如有矛盾,在处理第几个大小关系时检测出了矛盾?

    分析

      每输入一行就进行一次拓扑排序,暴力求解即可。

    代码如下

      1 #include <cmath>
      2 #include <ctime>
      3 #include <iostream>
      4 #include <string>
      5 #include <vector>
      6 #include <cstdio>
      7 #include <cstdlib>
      8 #include <cstring>
      9 #include <queue>
     10 #include <map>
     11 #include <set>
     12 #include <algorithm>
     13 #include <cctype>
     14 #include <stack>
     15 #include <deque>
     16 #include <list>
     17 #include <sstream>
     18 #include <cassert>
     19 using namespace std;
     20  
     21 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
     22 #define Rep(i,n) for (int i = 0; i < (n); ++i)
     23 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
     24 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
     25 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
     26 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
     27 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
     28 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
     29  
     30 #define pr(x) cout << #x << " = " << x << "  "
     31 #define prln(x) cout << #x << " = " << x << endl
     32  
     33 #define LOWBIT(x) ((x)&(-x))
     34  
     35 #define ALL(x) x.begin(),x.end()
     36 #define INS(x) inserter(x,x.begin())
     37 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
     38 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
     39 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
     40 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
     41  
     42 #define ms0(a) memset(a,0,sizeof(a))
     43 #define msI(a) memset(a,0x3f,sizeof(a))
     44 #define msM(a) memset(a,-1,sizeof(a))
     45 
     46 #define MP make_pair
     47 #define PB push_back
     48 #define ft first
     49 #define sd second
     50  
     51 template<typename T1, typename T2>
     52 istream &operator>>(istream &in, pair<T1, T2> &p) {
     53     in >> p.first >> p.second;
     54     return in;
     55 }
     56  
     57 template<typename T>
     58 istream &operator>>(istream &in, vector<T> &v) {
     59     for (auto &x: v)
     60         in >> x;
     61     return in;
     62 }
     63  
     64 template<typename T1, typename T2>
     65 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     66     out << "[" << p.first << ", " << p.second << "]" << "
    ";
     67     return out;
     68 }
     69 
     70 inline int gc(){
     71     static const int BUF = 1e7;
     72     static char buf[BUF], *bg = buf + BUF, *ed = bg;
     73     
     74     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
     75     return *bg++;
     76 } 
     77 
     78 inline int ri(){
     79     int x = 0, f = 1, c = gc();
     80     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
     81     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
     82     return x*f;
     83 }
     84 
     85 template<class T>
     86 inline string toString(T x) {
     87     ostringstream sout;
     88     sout << x;
     89     return sout.str();
     90 }
     91 
     92 inline int toInt(string s) {
     93     int v;
     94     istringstream sin(s);
     95     sin >> v;
     96     return v;
     97 }
     98 
     99 //min <= aim <= max
    100 template<typename T>
    101 inline bool BETWEEN(const T aim, const T min, const T max) {
    102     return min <= aim && aim <= max;
    103 }
    104  
    105 typedef long long LL;
    106 typedef unsigned long long uLL;
    107 typedef pair< double, double > PDD;
    108 typedef pair< int, int > PII;
    109 typedef pair< int, PII > PIPII;
    110 typedef pair< string, int > PSI;
    111 typedef pair< int, PSI > PIPSI;
    112 typedef set< int > SI;
    113 typedef set< PII > SPII;
    114 typedef vector< int > VI;
    115 typedef vector< char > VC;
    116 typedef vector< double > VD;
    117 typedef vector< VI > VVI;
    118 typedef vector< SI > VSI;
    119 typedef vector< PII > VPII;
    120 typedef map< int, int > MII;
    121 typedef map< LL, int > MLLI;
    122 typedef map< int, string > MIS;
    123 typedef map< int, PII > MIPII;
    124 typedef map< PII, int > MPIII;
    125 typedef map< string, int > MSI;
    126 typedef map< string, string > MSS;
    127 typedef map< PII, string > MPIIS;
    128 typedef map< PII, PII > MPIIPII;
    129 typedef multimap< int, int > MMII;
    130 typedef multimap< string, int > MMSI;
    131 //typedef unordered_map< int, int > uMII;
    132 typedef pair< LL, LL > PLL;
    133 typedef vector< LL > VL;
    134 typedef vector< VL > VVL;
    135 typedef priority_queue< int > PQIMax;
    136 typedef priority_queue< int, VI, greater< int > > PQIMin;
    137 const double EPS = 1e-8;
    138 const LL inf = 0x3fffffff;
    139 const LL infLL = 0x3fffffffffffffffLL;
    140 const LL mod = 1e9 + 7;
    141 const int maxN = 1e2 + 7;
    142 const LL ONE = 1;
    143 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
    144 const LL oddBits = 0x5555555555555555;
    145 
    146 struct Edge{
    147     int from, to;
    148 };
    149 
    150 istream& operator>> (istream& in, Edge &x) {
    151     in >> x.from >> x.to;
    152     return in;
    153 }
    154 
    155 struct Vertex{
    156     int in;
    157     VI next;
    158     
    159     void clear() {
    160         in = 0;
    161         next.clear();
    162     }
    163 };
    164 
    165 int N, M, ans, type;
    166 Vertex V[27];
    167 vector< Edge > E;
    168 string topo; // 存拓扑序列 
    169 
    170 void addEdge(Edge &x) {
    171     V[x.from].next.PB(E.size());
    172     ++V[x.to].in;
    173     E.PB(x);
    174 }
    175 
    176 // 返回 1 表示能确定顺序,0 表示不能确定,-1表示矛盾 
    177 int TopoSort() {
    178     stack< int > sk;
    179     int ind[27];
    180     topo.clear();
    181     
    182     For(i, 1, N) {
    183         if(!V[i].in) sk.push(i);
    184         ind[i] = V[i].in;
    185     }
    186     
    187     bool check = true; // check大小关系是否都确定了 
    188     while(!sk.empty()) {
    189         if(sk.size() > 1) check = false;
    190         int tmp = sk.top(); sk.pop();
    191         topo.PB(tmp + 'A' - 1);
    192         
    193         Rep(i, V[tmp].next.size()) {
    194             Edge &e = E[V[tmp].next[i]];
    195             
    196             --ind[e.to];
    197             if(!ind[e.to]) sk.push(e.to);
    198         }
    199     }
    200     
    201     if(topo.size() < N) return -1;
    202     if(!check) return 0;
    203     return 1;
    204 }
    205 
    206 int main(){
    207     //freopen("MyOutput.txt","w",stdout);
    208     //freopen("input.txt","r",stdin);
    209     //INIT();
    210     while(cin >> N >> M && N) {
    211         bool flag = false; 
    212         type = 0;
    213         E.clear();
    214         For(i, 1, N) V[i].clear();
    215                 
    216         For(i, 1, M) {
    217             string tmp;
    218             Edge t;
    219             
    220             cin >> tmp;
    221             if(flag) continue;
    222             
    223             t.from = tmp[0] - 'A' + 1;
    224             t.to = tmp[2] - 'A' + 1;
    225             if(tmp[1] == '>') swap(t.from, t.to);
    226             addEdge(t);
    227             
    228             int tt = TopoSort();
    229             if(tt == 1 || tt == -1) {
    230                 type = tt;
    231                 ans = i;
    232                 flag = true;
    233             }
    234         }
    235         
    236         if(type == -1) printf("Inconsistency found after %d relations.
    ", ans);
    237         else if(type == 1) printf("Sorted sequence determined after %d relations: %s.
    ", ans, topo.c_str());
    238         else printf("Sorted sequence cannot be determined.
    ", ans);
    239     }
    240     return 0;
    241 }
    View Code
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  • 原文地址:https://www.cnblogs.com/zaq19970105/p/11307830.html
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