题目传送门
1 /* 2 题意：几个判断，车上的人不能 <0 或 > C；车上初始和结束都不能有人在 (为0)；车上满员时才有等候的人 3 水题：难点在于读懂题目意思，状态不佳，一直没搞懂意思，在这题卡住了 4 主要要理解have to wait 的意思 5 */ 6 #include <cstdio> 7 #include <iostream> 8 #include <algorithm> 9 #include <cstring> 10 #include <string> 11 #include <cmath> 12 #include <set> 13 #include <map> 14 #include <queue> 15 using namespace std; 16 17 const int MAXN = 1e2 + 10; 18 const int INF = 0x3f3f3f3f; 19 int l[MAXN], e[MAXN], w[MAXN]; 20 21 int main(void) //Gym 100502K Train Passengers 22 { 23 //freopen ("K.in", "r", stdin); 24 25 int C, n; 26 27 while (scanf ("%d%d", &C, &n) == 2) 28 { 29 int i; 30 for (i=1; i<=n; ++i) 31 { 32 scanf ("%d%d%d", &l[i], &e[i], &w[i]); 33 } 34 long long sum = 0; 35 for (i=1; i<=n; ++i) 36 { 37 if (i == n && w[i] > 0) {puts ("impossible"); break;}; 38 sum -= l[i]; 39 if (sum < 0 || sum > C) {puts ("impossible"); break;}; 40 sum += e[i]; 41 if (sum < 0 || sum > C) {puts ("impossible"); break;}; 42 if (w[i] > 0 && sum < C) {puts ("impossible"); break;}; //当时没理解题意，写不出来:) 43 } 44 45 if (i > n) 46 { 47 if (sum != 0) puts ("impossible"); 48 else puts ("possible"); 49 } 50 } 51 52 return 0; 53 } 54 55 /* 56 impossible 57 possible 58 */
