UVALive - 4621 Cav

来源：互联网发布：linux和windows 编辑：程序博客网时间：2024/05/24 23:14

题意：在长n的地方，求能完全能放下的有的面积

思路：先从左到右，然后在从右到左统计每个位置能有的高度，最后再相加就是了

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 1000010;const int INF = 0x3f3f3f3f;int a[MAXN],b[MAXN];int n,h[MAXN];int main(){    int t;    scanf("%d",&t);    while (t--){        scanf("%d",&n);        for (int i = 1; i <= n; i++)            scanf("%d",&a[i]);        for (int i = 1; i <= n; i++)            scanf("%d",&b[i]);        a[0] = b[0] = INF;        a[n+1] = b[n+1] = INF;        int ah = INF;        for (int i = 1; i <= n; i++){            if (a[i-1] > ah)                ah = a[i-1];            if (a[i] > ah)                ah = a[i];            if (b[i] < ah)                ah = b[i];            h[i] = ah;        }        ah = INF;        for (int i = n; i >= 1; i--){            if (a[i+1] > ah)                ah = a[i+1];            if (a[i] > ah)                ah = a[i];            if (b[i] < ah)                ah = b[i];            if (ah < h[i])                h[i] = ah;            ah = h[i];        }        int ans = 0;        for (int i = 1; i <= n; i++)            ans += h[i] - a[i];        printf("%d\n",ans);    }    return 0;}

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