floyd + 01分数规划.
题中要求的是比率最大，那么自然就想到01分数规划。对于在哪几个城镇买卖商品，可以用O(n2 * k)贪心预处理。路程并不是两点间的距离，而是最短路，所以floyd先跑一遍。
因为答案下取整，所以整数二分就行。
此题卡dfs版spfa，只能用bfs版过……
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const ll INF = 1e14; 19 const db eps = 1e-8; 20 const int maxn = 105; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m, q; 38 ll G[maxn][maxn], v[maxn][maxn], tra[maxn][maxn * 10][2]; 39 40 ll dis[maxn]; 41 bool vis[maxn]; 42 int cnt[maxn]; 43 /*bool spfa(int now, ll x) 44 { 45 vis[now] = 1; 46 for(int i = 1; i <= n; ++i) 47 { 48 if(now != i && G[now][i] != INF && dis[i] >= dis[now] + x * G[now][i] - v[now][i]) 49 { 50 dis[i] = dis[now] + x * G[now][i] - v[now][i]; 51 if(vis[i]) return 1; 52 if(spfa(i, x)) return 1; 53 } 54 } 55 vis[now] = 0; 56 return 0; 57 }*/ 58 59 bool spfa(ll x) 60 { 61 queue<int> q; q.push(0); 62 while(!q.empty()) 63 { 64 int now = q.front(); q.pop(); vis[now] = 0; 65 for(int i = 1; i <= n; ++i) 66 if(now != i && G[now][i] != INF && dis[i] >= dis[now] + x * G[now][i] - v[now][i]) 67 //一定是大于等于,因为可能有零环！ 68 { 69 dis[i] = dis[now] + x * G[now][i] - v[now][i]; 70 if(!vis[i]) 71 { 72 vis[i] = 1, q.push(i); 73 if(++cnt[i] >= n) return 1; 74 } 75 } 76 } 77 return 0; 78 } 79 80 bool judge(ll x) 81 { 82 for(int i = 1; i <= n; ++i) dis[i] = INF; 83 Mem(vis, 0); Mem(cnt, 0); 84 dis[0] = 0; 85 return spfa(x); 86 } 87 88 int main() 89 { 90 n = read(); m = read(); q = read(); 91 for(int i = 1; i <= n; ++i) 92 for(int j = 1; j <= q; ++j) tra[i][j][0] = read(), tra[i][j][1] = read(); 93 for(int i = 1; i <= n; ++i) 94 for(int j = 1; j <= n; ++j) G[i][j] = INF; 95 for(int i = 1; i <= m; ++i) 96 { 97 int x = read(), y = read(); 98 ll w = read(); 99 G[x][y] = min(G[x][y], w); 100 } 101 for(int i = 1; i <= n; ++i) G[i][i] = 0; 102 for(int i = 1; i <= n; ++i) 103 for(int j = 1; j <= n; ++j) 104 { 105 for(int k = 1; k <= q; ++k) 106 if(tra[i][k][0] != -1 && tra[j][k][1] != -1) 107 v[i][j] = max(v[i][j], tra[j][k][1] - tra[i][k][0]); 108 } 109 for(int k = 1; k <= n; ++k) 110 for(int i = 1; i <= n; ++i) 111 for(int j = 1; j <= n; ++j) 112 G[i][j] = min(G[i][j], G[i][k] + G[k][j]); 113 for(int i = 1; i <= n; ++i) G[0][i] = 0; 114 ll L = 0, R = 1e9; 115 while(L < R) 116 { 117 ll mid = (L + R + 1) >> 1; 118 if(judge(mid)) L = mid; 119 else R = mid - 1; 120 } 121 write(L), enter; 122 return 0; 123 }