ZOJ 2699 Police Cities

Police Cities

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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Once upon the time there lived a king and he had a big kingdom. And there were n cities in his kingdom and some of them were connected by the roads. And the roads were all one-way because it would be dangerous if two carriages riding in opposite directions met on a road.

And once the king decided that he would like to establish police in his country and ordered to build police stations in some cities. But since his finances are limited, he would only like build police stations in k different cities. He would like to build them in such a way, that the following conditions were satisfied:

• it is possible to get by the roads from each city to some city with the police station;
• it is possible to get by the roads to each city from some city with the police station.

Now the king wants to know how many different ways are there to do so. Help him to find the answer to this question.

Input

There are mutilple cases in the input file.

The first line of each case contains n , m and k --- the number of cities and roads in the kingdom, and the number of police stations to build, respectively (1 <= n <= 100 , 0 <= m <= 20,000 , 1 <= k <= n ). The following m lines contain two city numbers each and describe roads, remember that it is only possible to travel along roads in one direction --- from the first city to the second one. Two cities may be connected by more than one road.

There is an empty line after each case.

Output

Output the only integer number --- the number of ways to fulfil king's request.

There should be an empty line after each case.

Sample Input

6 7 3
1 2
2 3
3 1
3 4
4 5
5 6
6 5

Sample Output

15

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Source: Andrew Stankevich's Contest #9

解题：强连通分量缩点后进行dp

首先是入度为0或者出度为0的点必须安排那个警察，dp[i][j]表示前i个入度或者出度为0的强连通分量安排了j个警察

那么有转移方程 $dp[i][j] = \sum_{k = 0}^{k < j} dp[i-1][k]*c[x][j-k]$ 这些分量是必须至少要安置一个警司的，x表示该分量内点的数量

至于其余的，至少安放0个

 #include<bits/stdc++.h>
 
 using namespace std;
 
 #define MAXN 100
 struct HP {
     int len,s[MAXN];
     HP() {
         memset(s,,sizeof(s));
         len=;
     }
     HP operator =(const char *num) { //字符串赋值
         len=strlen(num);
         for(int i=; i<len; i++) s[i]=num[len-i-]-'';
     }
 
     HP operator =(int num) { //int 赋值
         char s[MAXN];
         sprintf(s,"%d",num);
         *this=s;
         return *this;
     }
 
     HP(int num) {
         *this=num;
     }
 
     HP(const char*num) {
         *this=num;
     }
 
     string str()const { //转化成string
         string res="";
         for(int i=; i<len; i++) res=(char)(s[i]+'')+res;
         if(res=="") res="";
         return res;
     }
 
     HP operator +(const HP& b) const {
         HP c;
         c.len=;
         for(int i=,g=; g||i<max(len,b.len); i++) {
             int x=g;
             if(i<len) x+=s[i];
             if(i<b.len) x+=b.s[i];
             c.s[c.len++]=x%;
             g=x/;
         }
         return c;
     }
     void clean() {
         while(len >  && !s[len-]) len--;
     }
 
     HP operator *(const HP& b) {
         HP c;
         c.len=len+b.len;
         for(int i=; i<len; i++)
             for(int j=; j<b.len; j++)
                 c.s[i+j]+=s[i]*b.s[j];
         for(int i=; i<c.len-; i++) {
             c.s[i+]+=c.s[i]/;
             c.s[i]%=;
         }
         c.clean();
         return c;
     }
 
     HP operator - (const HP& b) {
         HP c;
         c.len = ;
         for(int i=,g=; i<len; i++) {
             int x=s[i]-g;
             if(i<b.len) x-=b.s[i];
             if(x>=) g=;
             else {
                 g=;
                 x+=;
             }
             c.s[c.len++]=x;
         }
         c.clean();
         return c;
     }
     HP operator / (const HP &b) {
         HP c, f = ;
         for(int i = len-; i >= ; i--) {
             f = f*;
             f.s[] = s[i];
             while(f>=b) {
                 f =f-b;
                 c.s[i]++;
             }
         }
         c.len = len;
         c.clean();
         return c;
     }
     HP operator % (const HP &b) {
         HP r = *this / b;
         r = *this - r*b;
         return r;
     }
 
     HP operator /= (const HP &b) {
         *this  = *this / b;
         return *this;
     }
 
     HP operator %= (const HP &b) {
         *this = *this % b;
         return *this;
     }
 
     bool operator < (const HP& b) const {
         if(len != b.len) return len < b.len;
         for(int i = len-; i >= ; i--)
             if(s[i] != b.s[i]) return s[i] < b.s[i];
         return false;
     }
 
     bool operator > (const HP& b) const {
         return b < *this;
     }
 
     bool operator <= (const HP& b) {
         return !(b < *this);
     }
 
     bool operator == (const HP& b) {
         return !(b < *this) && !(*this < b);
     }
     bool operator != (const HP &b) {
         return !(*this == b);
     }
     HP operator += (const HP& b) {
         *this = *this + b;
         return *this;
     }
     bool operator >= (const HP &b) {
         return *this > b || *this == b;
     }
 
 };
 
 istream& operator >>(istream &in, HP& x) {
     string s;
     in >> s;
     x = s.c_str();
     return in;
 }
 
 ostream& operator <<(ostream &out, const HP& x) {
     out << x.str();
     return out;
 }
 const int maxn = ;
 HP dp[maxn][maxn],c[maxn][maxn];
 void calc() {
     for(int i = ; i < maxn; ++i) {
         c[i][] = c[i][i] = ;
         for(int j = ; j < i; ++j)
             c[i][j] = c[i-][j-] + c[i-][j];
     }
 }
 int low[maxn],dfn[maxn],belong[maxn],bcc_size[maxn],bcc,clk;
 vector<int>g[maxn],x,y;
 stack<int>stk;
 int n,m,K,ind[maxn],outd[maxn];
 bool instack[maxn];
 void init() {
     for(int i = ; i < maxn; ++i) {
         dfn[i] = belong[i] = bcc_size[i] = ;
         g[i].clear();
         instack[i] = false;
         ind[i] = outd[i] = ;
     }
     bcc = clk = ;
     while(!stk.empty()) stk.pop();
     x.clear();
     y.clear();
 }
 void tarjan(int u) {
     dfn[u] = low[u] = ++clk;
     stk.push(u);
     instack[u] = true;
     for(int i = g[u].size() - ; i >= ; --i) {
         if(!dfn[g[u][i]]) {
             tarjan(g[u][i]);
             low[u] = min(low[u],low[g[u][i]]);
         } else if(instack[g[u][i]]) low[u] = min(low[u],dfn[g[u][i]]);
     }
     if(low[u] == dfn[u]) {
         bcc++;
         int v;
         do {
             v = stk.top();
             stk.pop();
             belong[v] = bcc;
             bcc_size[bcc]++;
             instack[v] = false;
         } while(v != u);
     }
 }
 int main() {
     calc();
     int u,v;
     while(~scanf("%d%d%d",&n,&m,&K)) {
         init();
         memset(dp,,sizeof dp);
         for(int i = ; i < m; ++i) {
             scanf("%d%d",&u,&v);
             g[u].push_back(v);
         }
         dp[][] = ;
         for(int i = ; i <= n; ++i)
             if(!dfn[i]) tarjan(i);
         for(int i = ; i <= n; ++i) {
             for(int j = g[i].size()-; j >= ; --j) {
                 if(belong[i] == belong[g[i][j]]) continue;
                 ind[belong[g[i][j]]]++;
                 outd[belong[i]]++;
             }
         }
         for(int i = ; i <= bcc; ++i)
             if(!ind[i] || !outd[i]) x.push_back(i);
             else y.push_back(i);
         for(int i = ; i <= x.size(); ++i)
             for(int j = ; j <= K; ++j)
                 for(int k = j-; k >=  && j - k <= bcc_size[x[i-]]; --k)
                     dp[i][j] += dp[i-][k]*c[bcc_size[x[i-]]][j - k];
         for(int i = x.size()+,t = ; i <= bcc; ++i,++t)
             for(int j = x.size(); j <= K; ++j)
                 for(int k = x.size(); k <= j; ++k)
                     dp[i][j] += dp[i-][k]*c[bcc_size[y[t]]][j - k];
         cout<<dp[bcc][K]<<endl<<endl;
     }
     return ;
 }