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题意:找出1到T的所有路径;
坑点:一开始以为是到终点,读错了题意,没测试第二个样例,结果WA了4遍,坑大了;
1 #include <iostream> 2 #include <cmath> 3 #include <cstdio> 4 #include <cstring> 5 #include <cstdlib> 6 #include <sstream> 7 #include <algorithm> 8 #define Max 2147483647 9 #define INF 0x7fffffff 10 #define N 90000+2 11 #define M 40000 12 #define ll long long 13 #define mem(a,b) memset(a,b,sizeof(a)) 14 #define repu(i, a, b) for(int i = (a); i < (b); i++) 15 #define repd(i, a, b) for(int i = (a-1); i >= (b); i--) 16 const double PI=-acos(-1.0); 17 using namespace std; 18 const int maxn = 999999; 19 int t = 0, v, sum, T; 20 int rec[30],vis[30]; 21 int g[30][30] ,floyd[30][30]; 22 void dfs(int x,int n) 23 { 24 if(x==T)///到终点输出 25 { 26 printf("%d" ,1); 27 repu(i,1,n-1) 28 printf(" %d",rec[i]); 29 printf(" %d\n",T); 30 sum++; 31 return ; 32 } 33 repu(i,1,v+1) 34 { 35 if(!vis[i]&&g[x][i]==1&&floyd[T][i]!=maxn) 36 {///i点能到达终点+x,i通路+未访问过 37 rec[n]=i; 38 vis[i]=1; 39 dfs(i,n+1); 40 vis[i]=0; 41 } 42 } 43 } 44 int main() 45 { 46 int x, y, cas = 1; 47 while (~scanf("%d", &T)) 48 { 49 v = 0; 50 repu(i,1,22) 51 repu(j,1,22) 52 g[i][j] = floyd[i][j] = maxn; 53 while(scanf("%d%d", &x, &y) && (x || y)) 54 { 55 g[x][y] = g[y][x] = 1;///无向图 56 floyd[x][y] = floyd[y][x] = 1; 57 v = max(v,max(x,y));///找最大点 58 } 59 for(int k = 1; k <= v; k++)///弗洛伊德算法找最短路,判断两点之间是否是通路 60 for (int i = 1; i <= v; i++) 61 for (int j = 1; j <= v; j++) 62 floyd[i][j] = min(floyd[i][j],floyd[i][k] + floyd[k][j]); 63 vis[1] = 1; 64 sum = 0; 65 printf("CASE %d:\n", cas++); 66 dfs(1, 1);///从第一组数找起 67 printf("There are %d routes from the firestation to streetcorner %d.\n", sum, T); 68 } 69 return 0; 70 }
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原文地址:http://www.cnblogs.com/ACMERY/p/4520598.html