題意：給出l和r，然后從l坐標到r坐標每隔兩個位置有一個檔板，給出擋板的高度，然后想（-1， 1）中間加水，問什么時候會溢出。

分析：只看一段區間的最大板塊

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for (int i = q; y[i] <= L; i++) {k += tmp; tmp = max(tmp, y[i+1]);}

這種方法很精妙，我想復雜了。?

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 1005;int l, r, x[N], y[N];int L, R, idl, idr;void init() {R = L = 0;for (int i = l; i <= r; i += 2) {if (i < 0) {scanf("%d", &x[(-i)/2]);if (L <= x[(-i)/2]) {L = x[(-i)/2]; idl = (-i)/2;}} else {scanf("%d", &y[i/2]);if (R < y[i/2]) {R = y[i/2]; idr = i/2;}}}}int del(int a, int b) {if (a <= b) return 2 * a;else return a + b;}int solve() {l = (-l) / 2; r = r / 2;if (R == L) {int k = 0, t = 0;int tmp = x[l];for (int i = l; i > idl; i--) {k += tmp; tmp = max(tmp, x[i-1]);}tmp = y[r];for (int i = r; i > idr; i--) {t += tmp; tmp = max(tmp, y[i-1]);}return (idl + idr + 1) * R * 2 + min(k, t) * 2 * 2;} else {int T = min(R, L);int p = 0, q = 0, k = 0, t = 0;while (p < l && x[p] < T) p++;while (q < r && y[q] < T) q++;if (R > L) {int tmp = y[q];for (int i = q; y[i] <= L; i++) {k += tmp; tmp = max(tmp, y[i+1]);}tmp = x[l];for (int i = l; i > p; i--) {t += tmp; tmp = max(tmp, x[i-1]);}} else {int tmp = x[p];for (int i = p; x[i] <= R; i++) {k += tmp; tmp = max(tmp, x[i+1]);}tmp = y[r];for (int i = r; i > q; i--) {t += tmp; tmp = max(tmp, y[i-1]);}}int ans = t > k ? t + k : 2 * t;return ans * 2 + (p + q + 1) * T * 2;}}int main() {while (scanf("%d%d", &l, &r) == 2 && l && r) {init();printf("%d\n", solve());}return 0;}

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