BZOJ3709 [PA2014]Bohater

贪心水题= =

如果可以的话，我们发现最后答案是确定的，故只要将序列分为两部分：

（1）杀完能回血的按照消耗升序

（2）剩余按血药回血量降序

然后模拟一遍判断是否合法即可

 

 1 /**************************************************************
 2     Problem: 3709
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:728 ms
 7     Memory:3152 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12  
13 using namespace std;
14 typedef long long ll;
15 const int N = 100005;
16  
17 int n, t1, t2;
18 ll z;
19  
20 struct data {
21     int d, a, w;
22     data() {}
23     data(int _d, int _a, int _w) : d(_d), a(_a), w(_w) {}
24 }a[N], b[N];
25 inline bool cmp_a(const data &a, const data &b) {
26     return a.d < b.d;
27 }
28 inline bool cmp_b(const data &a, const data &b) {
29     return a.a > b.a;
30 }
31  
32 inline int read() {
33     int x = 0;
34     char ch = getchar();
35     while (ch < '0' || '9' < ch)
36         ch = getchar();
37     while ('0' <= ch && ch <= '9') {
38         x = x * 10 + ch - '0';
39         ch = getchar();
40     }
41     return x;
42 }
43  
44 int len = 0, pr[15];
45 inline void print(int x) {
46     if (x == 0) {
47         putchar('0'), putchar(' ');
48         return;
49     }
50     while (x)
51         pr[++len] = x % 10, x /= 10;
52     while (len)
53         putchar(pr[len--] + '0');
54     putchar(' ');
55 }
56  
57 int main() {
58     int i, x, y;
59     n = read(), z = read();
60     for (i = 1; i <= n; ++i) {
61         x = read(), y = read();
62         if (x <= y) a[++t1] = data(x, y, i);
63         else b[++t2] = data(x, y, i);
64     }
65     sort(a + 1, a + t1 + 1, cmp_a);
66     for (i = 1; i <= t1; ++i)
67         if (z <= a[i].d) {
68             puts("NIE");
69             return 0;
70         } else
71         z -= a[i].d - a[i].a;
72     sort(b + 1, b + t2 + 1, cmp_b);
73     for (i = 1; i <= t2; ++i)
74         if (z <= b[i].d) {
75             puts("NIE");
76             return 0;
77         } else
78         z -= b[i].d - b[i].a;
79     puts("TAK");
80     for (i = 1; i <= t1; ++i)
81         print(a[i].w);
82     for (i = 1; i <= t2; ++i)
83         print(b[i].w);
84     puts("");
85     return 0;
86 }
View Code

（p.s. Rank 5，话说Rank 2一看就知道是二逼读入优化的23333）

posted on 2014-11-24 12:37  Xs酱~  阅读(152)  评论(0)  编辑  收藏  举报