Poj 2594 treasure transition ation [transfer closure + minimum path overwrite]

Note:

There are n vertices, which tell you how many vertices can cover all vertices. [Note that vertices can overlap for example, 4-> 1-> 2 5-> 3]

 

Analysis:

Knowledge reserve:

Transfer closure: the so-called transfer, it can be understood that for node J, if I can reach K and K can reach J, then I can reach J. In this way, like floyed, it can process whether the specified two points can arrive

for(int k = 1; k <= n; k++) {    for(int i = 1; i <= n; i++) {        if(W[i][k]) {            for(int j = 1; j <= n; j++) {                W[i][j] = W[i][j] || (W[i][k] && W[k][j]);            }        }    }}

 

 

Because the points can be reused, the passing closure is used to process the arrival relationship between any two points.

You can find the minimum path overwrite.

 

Code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 using namespace std; 6  7 const int maxn = 505; 8 const int INF = 1000000000; 9 10 int n;11 int vis[maxn];12 int Link[maxn];13 vector<int> G[maxn];14 bool Find(int u) {15     for(int i = 0; i < G[u].size(); i++) {16         int v = G[u][i];17         if(!vis[v]) {18             vis[v] = 1;19             if(Link[v] == -1 || Find(Link[v])) {20                 Link[v] = u;21                 return true;22             }23         }24     }25     return false;26 }27 28 int solve() {29     memset(Link, -1, sizeof(Link));30     int cnt = 0;31     for(int i = 1; i <= n; i++) {32         if(G[i].size()) {33             memset(vis, 0, sizeof(vis));34             if(Find(i)) cnt++;35         }36     }37     return cnt;38 }39 40 int W[maxn][maxn];41 int main() {42     int m;43     int a, b;44     while(scanf("%d %d",&n, &m) && ( n + m) ) {45         for(int i = 1; i <= n; i++) G[i].clear();46         memset(W, 0, sizeof(W));47         while(m--) {48             scanf("%d %d",&a, &b);49             W[a][b] = 1;50         }51         for(int k = 1; k <= n; k++) {52             for(int i = 1; i <= n; i++) {53                 if(W[i][k]) {54                     for(int j = 1; j <= n; j++) {55                         W[i][j] = W[i][j] || (W[i][k] && W[k][j]);56                     }57                 }58             }59         }60         for(int i = 1; i <= n; i++) {61             for(int j = 1; j <= n; j++) {62                 if(W[i][j]) G[i].push_back(j);63             }64         }65         printf("%d\n",n - solve());66     }67     return 0;68 }

View code

 

Poj 2594 treasure transition ation [transfer closure + minimum path overwrite]