最短路 参考了Staingger的博客
感觉DP的状态记录还是有毛病。可以DFS寻找结果也。
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} #define MAXN 150 const int INF = 0x3f3f3f3f ; int dp[MAXN][MAXN]; bool vis[MAXN][MAXN]; int N,M,ans; void init() { scanf("%d%d",&N,&M); memset(dp,0x3f,sizeof(dp)); while (M--) { int u,v; scanf("%d%d",&u,&v); dp[u][v] = dp[v][u] = 1; } for (int k = 0; k < N; k++) for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j]); for (int i = 0; i < N; i++) dp[i][i] = 0; } int calcu(int s, int t) { int res[MAXN],cas = 0,cnt = 0; for (int k = 0; k < N; k++) if (dp[s][k] + dp[k][t] == dp[s][t]) res[cas++] = k; for (int i = 0; i < cas; i++) for (int j = i + 1; j < cas; j++) { int a = res[i] , b = res[j]; if(dp[a][b] == 1 && dp[s][a] != dp[s][b]) cnt++; } return cnt; } void slove() { for (int i = 0; i < N; i++) for (int j = i + 1; j < N; j++) { if (dp[i][j] != INF) ans = max(ans,calcu(i,j)); } } int main() { //freopen("sample.txt","r",stdin); int T,kase = 1; scanf("%d",&T); while (T--) { init(); ans = 0; slove(); printf("Case #%d: %d\n",kase++,ans); } return 0; }
UVA 10985 Rings'n'Ropes
时间: 2024-11-13 15:19:50