POJ 3658 Artificial Lake (单调栈)

题意：

析：利用单调栈，维护一个单调递增的栈，首先在最低的平台开始，每次向两边进行扩展，寻找两边最低的，然后不断更新宽度。

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>
#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 freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 1e5 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -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;
}
struct Node{
  int w, h, id;
};
Node a[maxn], sta[maxn];
LL ans[maxn];

int main(){
  while(scanf("%d", &n) == 1){
    int minh = INF;
    int id;
    for(int i = 1; i <= n; ++i){
      scanf("%d %d", &a[i].w, &a[i].h);
      if(minh > a[i].h){
        minh = a[i].h;
        id = i;
      }
      a[i].id = i;
//      printf("%d\n", a[i].w);
    }
    int top = 0;
    a[0] = a[n+1] = (Node){INF, INF, INF};
    sta[++top] = a[0];
    sta[++top] = a[id];
//    printf("%d\n", sta[top].w);
    int l = id;  int r = id;
    LL all = 0;
    for(int i = 1; i <= n; ++i){
      int tmp;
      if(a[l-1].h < a[r+1].h)  tmp = --l;
      else tmp = ++r;
      int add = 0;
      while(top && a[tmp].h > sta[top].h){
        sta[top].w += add;
        ans[sta[top].id] = all + sta[top].w;
        all += (LL)(min(a[tmp].h, sta[top-1].h) - sta[top].h) * sta[top].w;
        //printf("%lld\n", all);
        add = sta[top].w;
        --top;
      }
      a[tmp].w += add;
      sta[++top] = a[tmp];
    }
    for(int i = 1; i <= n; ++i)
      printf("%lld\n", ans[i]);
  }
  return 0;
}