题目链接
1 #include 2 #include <string> 3 #include 4 #include 5 #include 6 #include 7 using namespace std; 8 #define LL __int64 9 #define MOD 1000000007 10 #define INF 0xffffff 11 int dp[101][101],g[101][101],pre[101][101]; 12 int ans[1001]; 13 int main() 14 { 15 int n,m,minz,i,j,k,sv,ev,w,num,temp; 16 while(scanf("%d%d",&n,&m)!=EOF) 17 { 18 memset(pre,-1,sizeof(pre)); 19 for(i = 1; i <= n; i ++) 20 { 21 for(j = 1; j <= n; j ++) 22 dp[i][j] = g[i][j] = INF; 23 dp[i][i] = g[i][i] = 0; 24 } 25 for(i = 1; i <= m; i ++) 26 { 27 scanf("%d%d%d",&sv,&ev,&w); 28 g[sv][ev] = dp[sv][ev] = min(dp[sv][ev],w); 29 g[ev][sv] = dp[ev][sv] = min(dp[ev][sv],w); 30 pre[sv][ev] = ev; 31 pre[ev][sv] = sv; 32 } 33 minz = INF; 34 for(i = 1; i <= n; i ++)//求最小环 35 { 36 for(j = 1; j <= i-1; j ++) 37 for(k = j+1; k <= i-1; k ++) 38 { 39 if(minz > dp[j][k] + g[j][i] + g[i][k]) 40 { 41 minz = dp[j][k] + g[j][i] + g[i][k]; 42 num = 1; 43 ans[0] = i; 44 temp = j; 45 while(temp != -1) 46 { 47 ans[num++] = temp; 48 temp = pre[temp][k]; 49 } 50 } 51 } 52 for(j = 1; j <= n; j ++) 53 for(k = 1; k <= n; k ++) 54 { 55 if(dp[j][k] > dp[j][i] + dp[i][k]) 56 { 57 dp[j][k] = dp[j][i] + dp[i][k]; 58 pre[j][k] = pre[j][i]; 59 } 60 } 61 } 62 if(minz == INF) 63 printf("No solution.\n"); 64 else 65 { 66 for(i = 0;i < num;i ++) 67 { 68 if(i == 0) 69 printf("%d",ans[i]); 70 else 71 printf(" %d",ans[i]); 72 } 73 } 74 printf("\n"); 75 } 76 return 0; 77 }
转载于:https://www.cnblogs.com/naix-x/p/3247486.html
