Tsinsen A1120 拦截导弹

http://oj.tsinsen.com/A1120

分析：第一问是最长不降子序列，第二问是贪心，重复第一问的操作直到所有导弹都被拦截即可

代码：

#include "cstdio"
using namespace std;

int cnt, s[50050];

int Dp(const int Num[], int n) {
    int dp[50050], prev[50050], Out = 1, Last = 0;
    for (int i = 0; i < 50050; ++i) {
        prev[i] = i; dp[i] = 1;
    }
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < i; ++j)
            if (s[Num[j]] >= s[Num[i]] && dp[i] < dp[j] + 1) {
                dp[i] = dp[j] + 1; prev[i] = j;
                if (Out < dp[i]) Out = dp[Last = i];
            }
    while (prev[Last] != Last) {
        s[Num[Last]] = -1; Last = prev[Last];
    }
    s[Num[Last]] = -1;
    return Out;
}

int main() {
    while (~scanf("%d", &s[cnt++]) && getchar() != '\n');
    int Num[50050] = { 0 }, Out = 0;
    for (int i = 0; i < cnt; ++i)
        Num[i] = i;
    printf("%d\n", Dp(Num, cnt));
    for (; ; ) {
        Out += 1;
        int Num[50050] = { 0 }, n = 0;
        for (int i = 0; i < cnt; ++i) if (s[i] != -1)
            Num[n++] = i;
        if (n <= 0) break;
        Dp(Num, n);
    }
    printf("%d\n", Out);
    return 0;
}