POJ 2433 Landscaping (贪心)

题意：给定一个序列表示一群山，要你保留最多 K 个山峰，最少要削去多少体积和土。一个山峰是指一段连续的相等的区间，并且左边和右边只能比这个区间低，或者是边界。

析：贪心，每次都寻找体积最小的山峰，然后把它削去，每次削的是最小的，所以是满足贪心的，最后剩下的小于 K 个就可以了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 1000 + 10;
const int maxm = 20;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }
inline double readDouble(){ double x;  scanf("%lf", &x);  return x; }

int a[maxn];

int main(){
  int K;
  while(scanf("%d %d", &n, &K) == 2){
    for(int i = 1; i <= n; ++i)  scanf("%d", a + i); a[n+1] = 0;
    
    int ans = 0;
    while(true){
      int cnt = 0, L, R;  m = INF;
      for(int i = 1; i <= n; ++i)  if(a[i] > a[i+1]){
        int l = i, r = i;
        while(l >= 1 && a[l-1] <= a[l])  --l;
        while(r <= n && a[r+1] <= a[r])  ++r;
        if(l >= 1 || r <= n){
          ++cnt;
          int t = max(a[l], a[r]);
          int sum = 0;
          for(int j = l+1; j < r; ++j)  if(a[j] > t)  sum += a[j] - t;
          if(sum < m){ m = sum;  L = l; R = r; }
        }
        i = r;
      }
      if(cnt <= K)  break;
      ans += m;
      int t = max(a[L], a[R]);
      for(int i = L+1; i < R; ++i) if(a[i] > t)  a[i] = t;
    }
    printf("%d\n", ans);
  }
  return 0;
}