loj1017

题目链接：点击打开链接

题解：dp状态方程： dp[i][j] = max(dp[i-1][j-cont[j]]+cont[j],dp[i-1][j-1]);

cont[j]:以第j个点为上沿时能刷到的点。

代码

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#define ll long longusing namespace std;const int N = 120;int n,w,k;int a[N];int cont[N];int dp[N][N];int solve(){ll h;cont[1] = 1;for (ll i = 1; i < n; i++){h = a[i]-w;ll j;for(j = i - 1; ; j--){if (j<0)break;if (a[j] < h)break;}cont[i+1] = i-j;}cont[0] = 0;for (int i = 1; i <= n; i++)memset(dp, 0, sizeof(dp));for (int i = 1; i <= k; i++)for (int j = 1; j <= n; j++)dp[i][j] = max(dp[i - 1][j - cont[j]] + cont[j], dp[i][j - 1]);return dp[k][n];}int main(){int test;scanf("%d", &test);int x, y;for (int cas = 1; cas <= test; cas++){scanf("%d%d%d", &n, &w, &k);for (int i = 0; i < n; i++)scanf("%d%d", &x, &a[i]);sort(a, a + n);printf("Case %d: %d\n", cas, solve());}return 0;}