usaco1.4.2 Mother's Milk
2016-12-01 15:56
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一. 原题
Mother's MilkFarmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk.
Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.
Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.
PROGRAM NAME: milk3
INPUT FORMAT
A single line with the three integers A, B, and C.SAMPLE INPUT (file milk3.in)
8 9 10
OUTPUT FORMAT
A single line with a sorted list of all the possible amounts of milk that can be in bucket C when bucket A is empty.SAMPLE OUTPUT (file milk3.out)
1 2 8 9 10
SAMPLE INPUT (file milk3.in)
2 5 10
SAMPLE OUTPUT (file milk3.out)
5 6 7 8 9 10
二. 分析
有3个给定容量(a,b,c<=20)的桶,初始时只有3号桶为满,1,2号桶为空。一次倒水只能把桶倒空或者把目标桶倒满。问1号桶为空时,3号桶里可能的残余的水的体积。搜索空间很小,总共C(21, 2)个状态,dfs即可。三. 代码
运行结果:USER: Qi Shen [maxkibb3] TASK: milk3 LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.000 secs, 4188 KB] Test 2: TEST OK [0.000 secs, 4188 KB] Test 3: TEST OK [0.000 secs, 4188 KB] Test 4: TEST OK [0.000 secs, 4188 KB] Test 5: TEST OK [0.000 secs, 4188 KB] Test 6: TEST OK [0.000 secs, 4188 KB] Test 7: TEST OK [0.000 secs, 4188 KB] Test 8: TEST OK [0.000 secs, 4188 KB] Test 9: TEST OK [0.000 secs, 4188 KB] Test 10: TEST OK [0.000 secs, 4188 KB] All tests OK.YOUR PROGRAM ('milk3') WORKED FIRST TIME! That's fantastic
-- and a rare thing. Please accept these special automated
congratulations.
AC代码:
/* ID:maxkibb3 LANG:C++ PROG:milk3 */ #include<cstdio> int v[3]; int a[3]; bool vis[21][21][21]; int min(int a, int b) { return (a < b)?a:b; } void dfs(int x[3]) { if(vis[x[0]][x[1]][x[2]]) return; vis[x[0]][x[1]][x[2]] = true; for(int i = 0; i < 3; i++) { for(int j = 0; j < 3; j++) { if(i == j) continue; if(x[i] != 0 && x[j] != v[j]) { int amt = min(x[i], v[j] - x[j]); x[i] -= amt; x[j] += amt; dfs(x); x[i] += amt; x[j] -= amt; } } } } int main() { freopen("milk3.in", "r", stdin); freopen("milk3.out", "w", stdout); scanf("%d%d%d", &v[0], &v[1], &v[2]); a[0] = a[1] = 0; a[2] = v[2]; dfs(a); for(int i = 0; i < v[2]; i++) { for(int j = 0; j <= v[2]; j++) { if(vis[0][j][i]) { printf("%d ", i); break; } } } printf("%d\n", v[2]); return 0; }
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