uva1442 Cav

连通器
向左向右扫描两次即可
每一段有水的连通区域,高度必须相同,且不超过最低天花板高度
if(p[i] > level) level = p[i]; 被隔断,要上升(隔断后,之前的就不变了,之后的从p【i】开始)
if(s[i] < level) level = s[i];
h[i] = level;
左右分别扫描一次,可以满足条件,取从左向右的h[i]和从右向左的level的min
ans += min(h[i], level) - p[i];

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 1000000 + 5;
int n, p[maxn], s[maxn], h[maxn];
int main() {
  int T;
  scanf("%d", &T);
  while(T--) {
    scanf("%d", &n);
    for(int i = 0; i < n; i++) scanf("%d", &p[i]);
    for(int i = 0; i < n; i++) scanf("%d", &s[i]);

    int ans = 0, level = s[0];
    for(int i = 0; i < n; i++) {
      if(p[i] > level) level = p[i];
      if(s[i] < level) level = s[i];
      h[i] = level;
    }
    level = s[n-1];
    for(int i = n-1; i >= 0; i--) {
      if(p[i] > level) level = p[i];
      if(s[i] < level) level = s[i];
      ans += min(h[i], level) - p[i];
    }
    printf("%d\n", ans);
  }
  return 0;
}

 

posted @ 2018-10-05 19:03  Erio  阅读(209)  评论(0编辑  收藏  举报