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这么个简单的题目居然没有人题解。floyd中计算数目,同时注意重边。
1 /* 1706 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 #include <iterator> 20 #include <iomanip> 21 using namespace std; 22 //#pragma comment(linker,"/STACK:102400000,1024000") 23 24 #define sti set<int> 25 #define stpii set<pair<int, int> > 26 #define mpii map<int,int> 27 #define vi vector<int> 28 #define pii pair<int,int> 29 #define vpii vector<pair<int,int> > 30 #define rep(i, a, n) for (int i=a;i<n;++i) 31 #define per(i, a, n) for (int i=n-1;i>=a;--i) 32 #define clr clear 33 #define pb push_back 34 #define mp make_pair 35 #define fir first 36 #define sec second 37 #define all(x) (x).begin(),(x).end() 38 #define SZ(x) ((int)(x).size()) 39 #define lson l, mid, rt<<1 40 #define rson mid+1, r, rt<<1|1 41 42 const int INF = 0x3f3f3f3f; 43 const int maxn = 105; 44 int dis[maxn][maxn]; 45 int M[maxn][maxn]; 46 int cnt[maxn][maxn]; 47 int n, m; 48 49 void floyd() { 50 rep(k, 1, n+1) { 51 rep(i, 1, n+1) { 52 if (M[i][k]>=INF || i==k) 53 continue; 54 rep(j, 1, n+1) { 55 if (M[k][j]>=INF || k==j || i==j) 56 continue; 57 if (M[i][j] > M[i][k]+M[k][j]) { 58 M[i][j] = M[i][k] + M[k][j]; 59 cnt[i][j] = cnt[i][k] * cnt[k][j]; 60 } else if (M[i][j] == M[i][k]+M[k][j]) { 61 cnt[i][j] += cnt[i][k] * cnt[k][j]; 62 } 63 } 64 } 65 } 66 } 67 68 int main() { 69 ios::sync_with_stdio(false); 70 #ifndef ONLINE_JUDGE 71 freopen("data.in", "r", stdin); 72 freopen("data.out", "w", stdout); 73 #endif 74 75 int u, v, w; 76 int mx; 77 int ans; 78 79 while (scanf("%d %d", &n, &m)!=EOF) { 80 memset(M, 0x3f, sizeof(M)); 81 memset(cnt, 0, sizeof(cnt)); 82 rep(i, 0, m) { 83 scanf("%d %d %d", &u, &v, &w); 84 if (u == v) 85 continue; 86 if (M[u][v] > w) { 87 M[u][v] = M[v][u] = w; 88 cnt[u][v] = cnt[v][u] = 1; 89 } else if (M[u][v] == w) { 90 ++cnt[u][v]; 91 ++cnt[v][u]; 92 } 93 } 94 floyd(); 95 ans = 0; 96 mx = -1; 97 rep(i, 1, n+1) { 98 rep(j, 1, i) { 99 if (M[i][j]==INF || i==j) 100 continue; 101 if (mx < M[i][j]) { 102 mx = M[i][j]; 103 ans = cnt[i][j]; 104 } else if (mx == M[i][j]) { 105 ans += cnt[i][j]; 106 } 107 } 108 } 109 printf("%d %d\n", mx, ans); 110 } 111 112 #ifndef ONLINE_JUDGE 113 printf("time = %d.\n", (int)clock()); 114 #endif 115 116 return 0; 117 }
【HDOJ】1706 The diameter of graph
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原文地址:http://www.cnblogs.com/bombe1013/p/4985070.html