[CFgym101061C]Ramzi（贪心，双条件最短路）

题目链接：http://codeforces.com/gym/101061/problem/C

题意：一张图，图上的边有两种，一种是车道，一种是人行道。一个人要从A点到B点，可以坐车也可以走人行道。这个人希望在走最少的路的情况下尽可能早地到达Ｂ点（保证走路最少的清空下坐车时间最少），问要走多少路，一共花费多久。

pair<int, int>保存这个人需要走路的时间和共计的时间，读入更新图的时候需要判断仔细。利用pair自身比较运算符优先判断first元素可以直接跑floyd。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <cassert>
 24 #include <cstdio>
 25 #include <bitset>
 26 #include <vector>
 27 #include <deque>
 28 #include <queue>
 29 #include <stack>
 30 #include <ctime>
 31 #include <set>
 32 #include <map>
 33 #include <cmath>
 34 //#include <unordered_map>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%I64d", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onenum(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70
 71 pii operator+(pii A, pii B) {
 72     return pii(A.first + B.first, A.second + B.second);
 73 }
 74 const int maxn = 1010;
 75 int n, m;
 76 pii dp[maxn][maxn];
 77
 78 int main() {
 79 //    FRead();
 80     int T;
 81     int x, y, c, k;
 82     Rint(T);
 83     W(T) {
 84         Rint(n); Rint(m);
 85         Rep(i, n+5) {
 86             Rep(j, n+5) dp[i][j] = pii(9000000, 9000000);
 87             dp[i][i] = pii(0, 0);
 88         }
 89         Rep(i, m) {
 90             Rint(x); Rint(y); Rint(c); Rint(k);
 91             if(k == 1) {
 92                 if(dp[x][y].first > c) {
 93                     dp[x][y].first = min(dp[x][y].first, c);
 94                     dp[y][x].first = min(dp[y][x].first, c);
 95                     dp[x][y].second = min(dp[x][y].second, c);
 96                     dp[y][x].second = min(dp[y][x].second, c);
 97                 }
 98             }
 99             else {
100                 if(dp[x][y].first != 0) {
101                     dp[x][y].first = dp[y][x].first = 0;
102                     dp[x][y].second = dp[y][x].second = c;
103                 }
104                 else {
105                     dp[x][y].second = min(dp[x][y].second, c);
106                     dp[y][x].second = min(dp[y][x].second, c);
107                 }
108             }
109         }
110         For(k, 1, n+1) For(i, 1, n+1) For(j, 1, n+1)
111         dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
112         Rint(x); Rint(y);
113         if(dp[x][y].second >= 9000000) puts("-1");
114         else printf("%d %d\n", dp[x][y].first, dp[x][y].second);
115     }
116     RT 0;
117 }