FZU-2188 Across the river I
Time Limit: 3000MS |
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Memory Limit: 32768KB |
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64bit IO Format: %i64d &%i64u |
Submit Status Description one day, Xiaoming needs to transport x sheep and y wolves to the opposite river. The ship can accommodate n animals and xiaoming. Every time Xiao Ming boating, must have at least one animal to accompany him, otherwise he will feel bored, uneasy. Whether on board or on shore, the wolf will eat the sheep if it exceeds the sheep. Xiao Ming needs to send all the animals to the opposite, and no sheep to be eaten, at least how many times he can cross the river. Input has multiple sets of data, the first line of each group entered 3 integers want x, y, N (0≤x, y,n≤200) Output If all animals can be sent across the river, and no sheep die, an integer is exported: the fewest number of times. Otherwise output-1. Sample Input 3 3 233 33 3 Sample Output 11-1 Hint The first example Number of ships direction left bank right bank (Wolf sheep) 0:0 0 3 3 0 0 1:2 0 > 1 3 2 0 2:1 0 < 2 3 1 0 3:2 0 > 0 3 3 0 4:1 0 < 1 3 2 0 5:0 2 > 1 1 2 2 6:1 1 < 2 2 1 1 7:0 2 > 2 0 1 3 8:1 0 < 3 0 0 3 9:2 0 > 1 0 2 3 10:1 0 < 2 0 1 3 11:2 0 > 0 0 3 3 Source FOJ Prize Month-March 2015 |
#include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath
> #include <queue> using namespace std;
int x,y,n;
int vis[210][210];
struct node{int w,s;
int step;
}s,e;
int BFS () {memset (vis,0,sizeof (VIS));
queue<node>q;vis[0][0]=1; S.s=s.w=s.step=0;
Q.push (s); while (! Q.empty ()) {Node now=q.front ();
Q.pop ();
for (int i=0;i<=x-now.s;++i) {for (int. j=0;j<=y-now.w;++j) {if (i&&j>i) break;
if (i+j==0) continue;
if (i+j>n) break;
Node u;
U.s=now.s+i;u.w=now.w+j;
if (u.s<u.w&&u.s!=0) continue;
if (x-u.s<y-u.w&&x-u.s!=0) continue;
if (!VIS[U.S][U.W]) {u.step=now.step+1;
if (u.s==x&&u.w==y) return u.step;
for (int a=0;a<=u.s;++a) {for (int. b=0;b<=u.w;++b) {if (a&&a<b) break;
if (a+b==0) continue;
if (a+b>n) break;
Node v;
V.s=u.s-a;v.w=u.w-b; if (v.s<v.w&&v.s!=0) contInue;
if (x-v.s<y-v.w&&x-v.s!=0) continue;
if (VIS[V.S][V.W]) continue;
Vis[v.s][v.w]=1;
v.step=u.step+1;
Q.push (v);
}}}}} return-1;
} int main () {while (scanf ("%d%d%d", &x,&y,&n)!=eof) {printf ("%d\n", BFS ());
} return 0; }