Fzu 2150 Fire Game Technique BFS

Title: Fzu 2150 Fire Game

Give a m*n figure, ' # ' means lawn, '. ' Show the open space, and then you can choose to fire in any of the two lawn lattices, the fire every 1 s will spread to the surrounding four lattice, ask those two points to make burning all the lawn spend the least time.

Analysis: This topic is a bit difficult to think about, but given the small data range, we can violently enumerate the points where any lawn is located, and then press the two points into the queue to BFS to a minimum that satisfies the condition.

By the way, Fzu 2141 Sub-bipartite Graph's thinking, the game did not come out.

This topic to think of the complex, and then found others with greedy binary staining, each point of greed to choose with its point of color more than the opposite color, so it can be. It seems to be complicated at the time.

AC Code:

#include <iostream> #include <cstdio> #include <vector> #include <cstring> #include <queue
> Using namespace std;
typedef long Long LL;
const int N = 15;
int dx[5] = {0,0,1,-1};
int dy[5] = {1,-1,0,0};
    struct Node {int x,y;
int cnt;
};
Vector<node> v;
Char Mp[n][n];
int vis[n][n];
int n,m;
    int BFS (Node a,node b) {memset (vis,0,sizeof (VIS));
    Queue<node> Q;
    VIS[A.X][A.Y] = vis[b.x][b.y] = 1;
    a.cnt = 0,b.cnt = 0;
    Q.push (a), Q.push (b);
    int ans = 0x3f3f3f3f;
    int cas = 1;
        while (!q.empty ()) {a = Q.front ();
        Q.pop ();
        printf ("%d%d%d\n", a.x,a.y,a.cnt);
        ans = a.cnt;
            for (int i = 0;i<4;i++) {b.x = a.x + dx[i];
            B.Y = A.y + dy[i];
            b.cnt = a.cnt + 1;
            printf ("b:%d%d%d\n", b.x,b.y,b.cnt); if (b.x>0 && b.y>0 && b.x<=n && b.y<=m && vis[b.x][b.y]==0 && mp[b.x][b.y]== ' # ') {vis[b.x][b.y] = 1;
            Q.push (b);
}} return ans; void print () {for (int i=1;i<=n;i++) {for (int j = 1;j<=m;j++) {printf ("%c")
        MP[I][J]);
    } puts ("");
    int main () {//freopen ("Input.txt", "R", stdin);
    int T;
    scanf ("%d", &t);
        for (int cas=1;cas<=t;cas++) {v.clear ();
        scanf ("%d%d", &n,&m);
            for (int i=1;i<=n;i++) {getchar ();
                for (int j=1;j<=m;j++) {scanf ("%c", &mp[i][j]);
            if (mp[i][j]== ' # ') V.push_back ((Node) {i,j,0});
        int ans = 0x3f3f3f3f; for (int i=0;i<v.size (), i++) {for (int j=i;j<v.size (), j + +) {int tmp
                = BFS (V[i],v[j]);
                bool OK = true;
  for (int k = 1;k<=n;k++)              {for (int f = 1;f<=m;f++) {if (vis[k][f]
                            = = 0 && mp[k][f]== ' # ') {OK = false;
                        Break
                } if (ok==false) break;
                if (ok) {ans = min (ans,tmp);
        } printf ("Case%d:", CAs);
        if (ans = = 0x3f3f3f3f) puts ("-1");
    else printf ("%d\n", ans);
return 0; }