poj 1948 Triangular Pastures(01背包+暴力）

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题目链接：poj 1948 Triangular Pastures

题目大意：给出若干个木棍的长度，要求用这些木棍组成一个三角形，求最大面积，不能组成输出-1.

解题思路：dp[i][j]表示长度为i和j的边是否能同时被组成，用01背包去计算出所有的可组成情况，另一条边就用s（和）-i-j；然后就是枚举i和j，维护最大值。

#include <stdio.h>#include <string.h>#include <math.h>const int N = 805;int n, s, dp[N][N];void add (int l) {for (int i = 800; i >= 0; i--) {for (int j = 800; j >= 0; j--) {if (dp[i][j]) {if (i + l < N)dp[i+l][j] = 1;if (j + l < N)dp[i][j+l] = 1;}}}}void init () {s = 0;memset(dp, 0, sizeof(dp));dp[0][0] = 1;int l;for (int i = 0; i < n; i++) {scanf("%d", &l);s += l;add(l);}}int area (double x, double y, double z) {double p = (x*x - y*y + z*z)/(2*z);double h = sqrt(x*x - p*p);return h*z/2 * 100;}void solve () {int ans = -1;for (int i = 0; i < N; i++) {for (int j = 0; j < N; j++) {if (dp[i][j] == 0) continue;int x = s - i - j;if (i + j <= x || i + x <= j || j + x <= i) continue;int tmp = area(i, j, x);if (tmp > ans)ans = tmp;}}printf("%d\n", ans);}int main () {while (scanf("%d", &n) == 1 && n) {init ();solve ();}return 0;}

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