• HDU 5125 Magic Ball DP+树状数组


    由于只要找1~x 中的最大值,然后线段树又容易MLE,所以这里可以用树状数组搞。

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <set>
    #include <bitset>
    #include <queue>
    #include <stack>
    #include <string>
    #include <iostream>
    #include <cmath>
    #include <climits>
    
    using namespace std;
    const int maxn = 1005;
    
    class BIT {
    	int val[maxn << 2], n;
    	inline int lowbit(int x) {
    		return x & (-x);
    	}
    
    public:
    
    	void init(int __n) {
    		n = __n;
    		memset(val, 0, sizeof(val));
    	}
    	void addv(int pos, int v) {
    		while(pos <= n) {
    			val[pos] = max(val[pos], v);
    			pos += lowbit(pos);
    		}
    	}
    
    	int GetMax(int pos) {
    		int ret = 0;
    		while(pos >= 1) {
    			ret = max(ret, val[pos]);
    			pos -= lowbit(pos);
    		}
    		return ret;
    	}
    };
    
    BIT bit[maxn];
    int a[maxn], b[maxn], f[maxn][maxn][2];
    int n, m, num[maxn << 2], numcnt;
    
    inline int GetID(int Val) {
    	return lower_bound(num, num + numcnt, Val) - num + 1;
    }
    
    int main() {
    	int T; scanf("%d", &T);
    	for(int kase = 1; kase <= T; kase++) {
    		memset(f, 0, sizeof(f));
    		scanf("%d%d", &n, &m);
    		numcnt = 0;
    		for(int i = 1; i <= n; i++) {
    			scanf("%d%d", &a[i], &b[i]);
    			num[numcnt++] = a[i];
    			num[numcnt++] = b[i];
    		}
    		sort(num, num + numcnt);
    		numcnt = unique(num, num + numcnt) - num;
    		for(int i = 1; i <= n; i++) {
    			a[i] = GetID(a[i]);
    			b[i] = GetID(b[i]);
    		}
    		for(int i = 0; i <= m; i++) bit[i].init(numcnt);
    		int ans = 0;
    		for(int i = 1; i <= n; i++) {
    			for(int j = m; j >= 0; --j) {
    				int prev_max = bit[j].GetMax(a[i] - 1);
    				f[i][j][0] = prev_max + 1;
    				bit[j].addv(a[i], f[i][j][0]);
    				ans = max(ans, f[i][j][0]);
    				if(j == 0) continue;
    				prev_max = bit[j - 1].GetMax(b[i] - 1);
    				f[i][j][1] = prev_max + 1;
    				bit[j].addv(b[i], f[i][j][1]);
    				ans = max(ans, f[i][j][1]);
    			}
    		}
    		printf("%d
    ", ans);
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/4133337.html
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