uva208

题意:

从1出发,到达给的点有几种走法

所有的走法显然是需要回朔来求的,为避免超时,我们要剪枝

那么最重要的就是用到的flo,这个方法可以判断通路(即当前点i是否可达终点),如果不可达就剪枝

那么没达到终点,我们就把路径输出,路径存于一个数组中

代码:

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mmax = 0x3f3f3f3f;
int n, v;
int g[30][30];
int vis[30];
int sum;
int flo[30][30];
int rec[30];
void dfs(int cur ,int s) {
if(cur == n) {
printf("1");
for(int i=1; i<s-1; i++)
printf(" %d", rec[i]);
printf(" %d\n", n);
sum++;
return;
}
for(int i=1; i<=v; i++) {
if(flo
[i]!=mmax && !vis[i] &&g[cur][i]==1) {
rec[s] = i;
vis[i] = 1;
dfs(i, s+1);
vis[i] = 0;
}
}

}

int main() {
int cas = 1;
while(scanf("%d", &n) != EOF) {
v = 0;
int a, b;
for(int i=0; i<=21; i++)
for(int j=0; j<=21; j++)
g[i][j] = flo[i][j] = mmax;
while(scanf("%d%d", &a, &b) && (a||b)) {
g[a][b] = g[b][a] = 1;
flo[a][b] = flo[b][a] = 1;
v = max(max(a, b) ,v);
}
for(int k=1; k<=v; k++)
for(int i=1; i<=v; i++)
for(int j=1; j<=v; j++)
if(flo[i][j] > flo[i][k]+flo[k][j]) {
flo[i][j] = flo[i][k]+flo[k][j];

}
vis[1] = 1;
sum = 0;
printf("CASE %d:\n", cas++);
dfs(1, 1);
printf("There are %d routes from the firestation to streetcorner %d.\n", sum, n);
}

return 0;
}