例题1.1 勇者斗恶龙 UVa11292

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1.题目描述：点击打开链接

2.解题思路：本题用贪心法解决，事先对恶龙的直径，能力值从小到大排序，由于要试图杀掉所有的恶龙，因此枚举每只恶龙，用cur来表示当前骑士的下标，如果他恰好能杀掉恶龙，就用他，否则尝试下一个。如果cur与m相等时退出循环，若此时恶龙仍未被杀光，则无解，反之输出ans。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;const int N = 20000 + 10;int a[N], b[N];int main(){freopen("t.txt", "r", stdin);int n, m;while (cin >> n >> m && (n || m)){memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));for (int i = 0; i < n; i++)cin >> a[i];for (int i = 0; i < m; i++)cin >> b[i];sort(a, a + n);sort(b, b + m);int ans = 0, i;int cur = -1,flag = 1;for (i = 0; i < n; i++)//枚举恶龙{while (++cur < m)if (b[cur] >= a[i]) { ans += b[cur]; break; }if (cur == m)break;}if (i < n)flag = 0;if (!flag)printf("Loowater is doomed!\n");else printf("%d\n", ans);}return 0;}

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