题意很好理解的贪心题，然而却卡疯了的精度坑。
再次理解一下double小数运算时可能导致的精度问题，本题为避免该问题可以将小数乘以100化为整数进行比较，输出的时候再除以100就ok；
思路也很好想，数据也不大，直接贴代码吧。
1 #include <cstdio> 2 #include <iostream> 3 #include <sstream> 4 #include <cmath> 5 #include <cstring> 6 #include <cstdlib> 7 #include <string> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <queue> 12 #include <stack> 13 #include <algorithm> 14 using namespace std; 15 #define ll long long 16 #define _cle(m, a) memset(m, a, sizeof(m)) 17 #define repu(i, a, b) for(int i = a; i < b; i++) 18 #define repd(i, a, b) for(int i = b; i >= a; i--) 19 #define sfi(n) scanf("%d", &n) 20 #define pfi(n) printf("%d\n", n) 21 #define MAXN 100010 22 const int maxn = 16400; 23 struct Player 24 { 25 double a, b, c; 26 int max_; 27 set<int> s; 28 } p[maxn]; 29 int r[maxn], vis[maxn]; 30 int main() 31 { 32 int n; 33 int T = 1; 34 while(~scanf("%d", &n) && n) 35 { 36 for(int i = 0; i <= n; i++) 37 p[i].s.clear(); 38 for(int i = 1; i <= n; i++) 39 { 40 scanf("%lf%lf%lf", &p[i].a, &p[i].b, &p[i].c); 41 int a = (p[i].a+0.005)*100; //注意此处控制精度，其实就是为了四舍五入 42 int b = (p[i].b+0.005)*100; 43 int c = (p[i].c+0.005)*100; 44 45 p[i].s.insert(a); 46 p[i].s.insert(b); 47 p[i].s.insert(c); 48 p[i].s.insert(a+b); 49 p[i].s.insert(a+c); 50 p[i].s.insert(b+c); 51 p[i].s.insert(a+b+c); 52 p[i].s.insert(0); 53 p[i].max_ = a+b+c; 54 } 55 bool ok = true; 56 for(int i = 0; i < n; i++) 57 { 58 scanf("%d", &r[i]); 59 } 60 int ans = p[r[0]].max_; 61 for(int i = 1; i < n; i++) 62 { 63 int last = r[i-1], now = r[i]; 64 if(last > now) 65 { 66 if(*p[now].s.begin() >= ans) 67 { 68 ok = false; 69 break; 70 } 71 set<int>::iterator it = p[now].s.lower_bound(ans); 72 it--; 73 ans = *it; 74 } 75 else 76 { 77 if(*p[now].s.begin() > ans) 78 { 79 ok = false; 80 break; 81 } 82 set<int>::iterator it = p[now].s.lower_bound(ans); 83 if(it == p[now].s.end() || *it > ans) 84 { 85 --it; 86 ans = *it; 87 } 88 } 89 } 90 if(ok) 91 { 92 printf("Case %d: %.2lf\n", T++, (double)ans/100.00); 93 } 94 else 95 { 96 printf("Case %d: No solution\n", T++); 97 } 98 } 99 return 0; 100 }
