解題區/解題區 - UVa

UVa 1115 - Water Shortage

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

根據連通管原理，給定要灌入水的容積，請問最後水位的位置為何？

給定每個水槽的長寬高、以及離地面的高度。

Sample Input

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1
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11.0 7.0 5.0 1.0
15.0 6.0 4.0 1.0
 5.0 8.0 5.0 1.0
19.0 4.0 8.0 1.0
78.0

Sample Output

1
17.00

Solution

講到連通管原理，很容易想起帕斯卡原理 … 喵帕斯。

二分水位高度，計算容積與目標容積的差異。

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#include <stdio.h> 
#include <math.h>
#include <assert.h>
#include <algorithm>
using namespace std;
const int MAXN = 1024;
double LV[MAXN], H[MAXN], W[MAXN], D[MAXN], V;
void solve(int n, double V) {
    double l = 0, r = 0, mid;
    double sumV = 0;
    for (int i = 0; i < n; i++) {
        sumV += H[i] * W[i] * D[i];
        r = max(r, LV[i] + H[i]);
    }
    if (sumV < V) {
        puts("OVERFLOW");
        return ;
    }
    
    for (int it = 0; it < 100; it++) {
        mid = (l + r)/2;
        sumV = 0;
        for (int i = 0; i < n; i++) {
            if (mid < LV[i])	continue;
            sumV += W[i] * D[i] * min(H[i], mid - LV[i]);
        }
        if (sumV > V)
            r = mid;
        else
            l = mid;
    }
    printf("%.2lf\n", l);
}
int main() {
    int testcase, n;
    scanf("%d", &testcase);
    while (testcase--) {
        scanf("%d", &n);
        assert(n < MAXN);
        for (int i = 0; i < n; i++)
            scanf("%lf %lf %lf %lf", &LV[i], &H[i], &W[i], &D[i]);
        scanf("%lf", &V);
    
        solve(n, V);
        if (testcase)	puts("");
    }
    return 0;
}