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HDU 1052 Tian Ji -- The Horse Racing (贪心)
题意:田忌赛马,问你田忌最多能赢多少银子。
析:贪心,绝对贪心的题,贪心策略是:
1.如果田忌当前的最快的马能追上齐王的,那么就直接赢一局
2.如果田忌当前的最慢的马能追上齐王的,那么就直接赢一局
3.如果田忌当前的最慢的马不能超过齐王的,那么就输一局,并把齐王最快的干掉
通过以上策略,就是田忌赢的最多。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <unordered_map> #include <unordered_set> #define debug() puts("++++"); #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 5; const int mod = 2000; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn], b[maxn]; int main(){ while(scanf("%d", &n) == 1 && n){ for(int i = 0; i < n; ++i) scanf("%d", a+i); for(int i = 0; i < n; ++i) scanf("%d", b+i); sort(a, a + n, greater<int>()); sort(b, b + n, greater<int>()); int ans = 0; int fro1 = 0, fro2 = 0; int rear1 = n-1, rear2 = n-1; while(fro1 <= rear1){ if(a[fro1] > b[fro2]){ ans += 200; ++fro1; ++fro2; } else if(a[rear1] > b[rear2]){ ans += 200; --rear1; --rear2; } else if(a[rear1] <= b[rear2]){ if(a[rear1] < b[fro2]) ans -= 200; --rear1; ++fro2; } } printf("%d\n", ans); } return 0; } /* 3 15 12 9 16 13 9 5 7 8 9 12 15 7 8 9 13 16 4 1 2 4 5 2 3 3 4 */
HDU 1052 Tian Ji -- The Horse Racing (贪心)
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