1475 建设国家 DP
http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1475
这题转化过来就是,给定n个点,每个点都有一个过期时间,一个价值。现在安排他们成一直线,使得总和最大。
一开始就是贪心,这是一个很经典的贪心。
http://www.cnblogs.com/liuweimingcprogram/p/6358456.html
和这题差不多,然后这个比较特别,有一个地方是可以接两个,我就暴力枚举了。然后蜜汁wa
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> #include <bitset> const int maxn = 1e3 + 20; struct node { int val, cost; node(int a, int b) : val(a), cost(b) {} node() {} bool operator < (const struct node & rhs) const { if (val != rhs.val) return val > rhs.val; else return cost < rhs.cost; } }a[maxn]; int w[maxn]; bool del[maxn]; vector<struct node>vc; void work() { int n, H; cin >> n >> H; for (int i = 1; i <= n; ++i) { cin >> a[i].cost >> a[i].val; a[i].cost = H - a[i].cost; } sort(a + 1, a + 1 + n); int ans = 0; for (int i = 1; i <= n; ++i) { int t = a[i].cost; if (t > n) { ans += a[i].val; //必定可以,空位也少了一个了,就放去第n天 del[i] = true; continue; } while (t >= 1 && w[t]) { t--; } if (t) { w[t] = a[i].val; ans += a[i].val; del[i] = true; } } for (int i = 1; i <= n; ++i) { if (del[i]) continue; vc.push_back(a[i]); } // cout << w[1] << endl; // cout << w[2] << endl; // cout << w[3] << endl; // cout << w[4] << endl; // cout << ans << endl; int tans = 0; for (int i = 1; i <= n; ++i) { tans += w[i]; int tofind = -inf; if (w[i] == 0) continue; for (int j = 0; j < vc.size(); ++j) { if (vc[j].cost < i) continue; tofind = max(tofind, vc[j].val); } ans = max(ans, tans + tofind); } cout << ans << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif work(); return 0; }
看了题解后居然是dp
用dp[i][j]表示前i个点中,选出了j个点,的最大答案。
那么,要求解dp[i][j],则有dp[i][j] = dp[i - 1][j],就是不选第i个点。
如果要选,那么有两种情况,
1、把这个点作为结束点,也就是两个点放在一起。这需要这个点的保质期 >= j - 1即可。因为现在求解的是选了j个点,那么以前就是选了j - 1个点,排在一起,所以只需要保质期 >= j - 1,然后更新答案,不更新dp数组
2、接在后一排,那么需要保质期 >= j
特别地,一开始就过期的,就不能选。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> #include <bitset> const int maxn = 1e3 + 20; int dp[maxn][maxn]; struct node { int cost, val; bool operator < (const struct node & rhs) const { if (cost != rhs.cost) return cost < rhs.cost; else return val > rhs.val; } }a[maxn]; void work() { int n, h; cin >> n >> h; for (int i = 1; i <= n; ++i) { cin >> a[i].cost >> a[i].val; a[i].cost = h - a[i].cost; } sort(a + 1, a + 1 + n); // memset(dp, -0x3f, sizeof dp); dp[0][0] = 0; int ans = 0; for (int i = 1; i <= n; ++i) { if (a[i].cost == 0) continue; for (int j = 1; j <= i; ++j) { dp[i][j] = dp[i - 1][j]; if (a[i].cost >= j - 1) { ans = max(ans, dp[i - 1][j - 1] + a[i].val); } if (a[i].cost >= j) { dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + a[i].val); } } } cout << ans << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif work(); return 0; }
既然选择了远方,就要风雨兼程~
posted on 2017-02-13 19:26 stupid_one 阅读(198) 评论(0) 编辑 收藏 举报