250pt:
水题set处理。
500pt：
题意：
给你一个图，每条边关联的两点为朋友，题目要求假设x的金钱为y，则他的左右的朋友当中的钱数z，取值为y - d <= z <= y + d.求使得任意两点的最大金钱差值，若果是inf输出-1.
思路：
求任意两点的最短的的最大值即可，比赛时不知道哪地方写搓了，直接被系统样例给虐了，老师这么悲剧500有思路能写，老师不仔细哎..
floyd求任意两点的最短距离好写一些。
#include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define ll __int64 #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define lowbit(x) (x)&(-x) #define keyTree (chd[chd[root][1]][0]) #define Read() freopen("din.txt", "r", stdin) #define Write() freopen("dout.txt", "w", stdout); #define M 25000 #define N 207 using namespace std; const int inf = 0x7f7f7f7f; const int mod = 1000000007; int f[N][N]; int n; void floyd() { for (int k = 1; k <= n; ++k) { for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { if (f[i][k] != inf && f[k][j] != inf && f[i][j] > f[i][k] + f[k][j]) { f[i][j] = f[i][k] + f[k][j]; } } } } } class Egalitarianism { public: int maxDifference(vector <string> isF, int d) { n = isF.size(); for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) f[i][j] = inf; } for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { if (isF[i][j] == 'Y') { f[i + 1][j + 1] = 1; } } } floyd(); int Ma = 0; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { if (i == j) continue; Ma = max(Ma,f[i][j]); } } if (Ma == inf) return -1; else return Ma*d; } };