Travel time limits on canoes: MS | Memory limit: 65535 KB Difficulty: 2
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Describe
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For a canoe trip, canoes can be rented at the port, and there is no difference between them. A canoe can take up to two persons and the total weight of the passenger must not exceed the maximum capacity of the canoe. We want to minimize the cost of this activity, so find the minimum number of canoe bars that can accommodate all passengers. Now please write a program that reads the maximum capacity of the canoe, the number of passengers and the weight of each passenger. According to the rules given, calculate the minimum number of canoe bars that must be placed for all passengers and output the results.
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Input
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The
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first line enters S, which represents the number of groups of test data;
The first row of each group of data includes two integers w,n,80<=w<=200,1<=n<=300,w is the maximum carrying capacity of a canoe, n is the number of people;
The next set of data is the weight of each person (not larger than the ship's carrying capacity);
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Output
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The
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minimum number of canoes required for each group of people.
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Sample input
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385 65 84 85 80 84 8390 390 45 60100 550 50 90 40 60
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Sample output
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533
Can think of this to let the weight from small to large sort, let the smallest and biggest plus see can overload, if super let him and the second heavy people add together, if not super, let the number minus one, prove can save a boat!
is also a typical greedy algorithm
1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5 intw[ -];6 7 8 intMain ()9 {Ten intn,n,i,s; Onescanf"%d",&N); A while(n--) - { -scanf"%d%d",&s,&n); the for(i=0; i<n;i++) -scanf"%d",&w[i]); -Sort (w,w+n); - intk=n-1, q=N; + for(i=0;i<K;) - { + if(w[i]+w[k]<=s) A { atq--; -i++; -k--; - } - Else -k--;//Can not sit with the next person to compare in } -printf"%d\n", q); to } + return 0; -}
Travel on a canoe--nyoj topic 71