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最新推荐文章于 2019-11-11 22:00:58 发布

原创最新推荐文章于 2019-11-11 22:00:58 发布 · 252 阅读

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/*状态：dp[i][j], 表示人在第i秒在j个位置上所得到的最大馅饼数
状态转移: dp[i][j] += max(dp[i+1][j], dp[i+1][j-1], dp[i+1][j+1])*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define MAXCOL	11
#define MAXROW	100001

unsigned long long dp[MAXROW][MAXCOL];

unsigned long long dynamic_programming(int max_v)
{
	unsigned long long val = 0;
	for(int i = max_v-1; i >= 1; i --) {
		for(int j = 0; j < MAXCOL; j ++) {
			val = dp[i+1][j];
			if( j-1 >= 0 ) {
				val = max(val, dp[i+1][j-1]);
			}
			if( j+1 < MAXCOL ) {
				val = max(val, dp[i+1][j+1]);
			}
			dp[i][j] += val;
		}
	}
	return max(max(dp[1][4], dp[1][5]), dp[1][6]);
}

int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
	freopen("test.in", "r", stdin);
#endif
	int n, x, y, max_v;
	while( scanf("%d", &n) && n ) {
		max_v = 0;
		memset(dp, 0, sizeof(dp));
		for(int i = 0; i < n; i ++) {
			scanf("%d %d", &y, &x);
			dp[x][y] ++;
			max_v = max(max_v, x);
		}
		printf("%lld\n", dynamic_programming(max_v));
	}
	return 0;
}