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;
}