「题解」洛谷 UVA1619 感觉不错 Feel Good

题目

UVA1619 感觉不错 Feel Good

简化题意

找一段区间使得这段区间的最小值乘这段区间的元素和最大，在保证最大的前提下保证区间最短，在以上前提下保证左端点最小。

思路

单调栈，悬线法。

在最小值确定的情况下区间越长越好（除了最小值是 (0)）

Code

悬线法：

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#define M 100001

typedef long long ll;

ll ans, left, right, sum[M];
int n, l[M], r[M], a[M];

inline void read(int &T) {
    int x = 0;
    bool f = 0;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = !f;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    T = f ? -x : x;
}

int main() {
    int cnt = 0;
    while (1) {
        if (!cnt) scanf("%d", &n);
        ++cnt, ans = 0;
        memset(sum, 0, sizeof sum);
        for (int i = 1; i <= n; ++i) {
            read(a[i]);
            sum[i] = sum[i - 1] + a[i];
            l[i] = r[i] = i;
        }
        for (int i = 1; i <= n; ++i) {
            while (l[i] > 1 && a[i] <= a[l[i] - 1]) l[i] = l[l[i] - 1];
        }
        for (int i = n; i >= 1; --i) {
            while (r[i] < n && a[i] <= a[r[i] + 1]) r[i] = r[r[i] + 1];
        }
        for (int i = 1; i <= n; ++i) {
            if (a[i] == 0) l[i] = r[i] = i;
            if ((sum[r[i]] - sum[l[i] - 1]) * 1ll * a[i] > ans) {
                ans = (sum[r[i]] - sum[l[i] - 1]) * 1ll * a[i];
                right = r[i], left = l[i];
            }
            if ((sum[r[i]] - sum[l[i] - 1]) * 1ll * a[i] == ans) {
                if (r[i] - l[i] < right - left) {
                    right = r[i], left = l[i];
                }
            }
        }
        printf("%lld
", ans);
        printf("%lld %lld
", left, right);
        if (scanf("%d", &n) == EOF || !n) break;
        puts("");
    }
    return 0;
}

单调栈：

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#define M 100002

typedef long long ll;

ll ans, left, right, sum[M];
int n, top, s[M], w[M], a[M];

inline void read(int &T) {
    int x = 0;
    bool f = 0;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = !f;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    T = f ? -x : x;
}

int main() {
    int cnt = 0;
    while (1) {
        if (!cnt) scanf("%d", &n);
        ++cnt, ans = -1, left = 0, right = 0, top = 0;
        memset(sum, 0, sizeof sum);
        for (int i = 1; i <= n; ++i) {
            read(a[i]);
            sum[i] = sum[i - 1] + a[i];
        }
        a[n + 1] = 0, s[++top] = 1, w[top] = 1;
        for (int i = 2; i <= n + 1; ++i) {
            if (a[s[top]] < a[i]) {
                s[++top] = i;
                w[top] = 1;
            }
            else {
                int len = 0;
                while (top && a[s[top]] >= a[i]) {
                    len += w[top];
                    int r = i - 1, l = r + 1 - len;
                    if (a[s[top]] == 0) r = s[top], l = s[top];
                    if (1ll * (sum[r] - sum[l - 1]) * a[s[top]] > ans) {
                        ans = 1ll * (sum[r] - sum[l - 1]) * a[s[top]];
                        right = r, left = l;
                    }
                    if (1ll * (sum[r] - sum[l - 1]) * a[s[top]] == ans) {
                        if (l < left && len <= right - left + 1) {
                            right = r, left = l;
                        }
                    }
                    --top;
                }
                s[++top] = i, w[top] = len + 1;
            }
        }
        printf("%lld
", ans);
        printf("%lld %lld
", left, right);
        if (scanf("%d", &n) == EOF || !n) break;
        puts("");
    }
    return 0;
}