poj 3211 Washing Clothes 背包

       这题做得相当顺利，1A

       题意是求夫妇两人洗完衣服用的最小时间。先按衣颜色分类，对于每种颜色的衣服，最优解是都平分即V/2，背包容量为洗衣服的花费，用01背包来标记是否可以通过组合组成某个容量。若不能平均分，就从V/2开始搜索，最接近V/2的较大值就为两人洗完每种颜色衣服的最短时间，把各种颜色衣服的解加起来即为答案

 

#include <iostream>
#include <map>
#include <string>
#include <vector>
#include <cstring>
using namespace std;

const int MAX = 100*1001;
const int N = 101;

bool isvisit[MAX];
int cap[N];
map<string, int> cloth;
vector<int> group[N];

int m, n;

int find(int color)
{
	int size = group[color].size();
	int sum = 0;
	memset(isvisit, false, sizeof(isvisit));
	for (int i = 0; i < size; i++)
		sum += cap[ group[color][i] ];

	isvisit[0] = true;
	for (int i = 0; i < size; i++)
		for (int v = sum; v >= cap[ group[color][i] ]; v--)
			 if (!isvisit[v] && isvisit[v - cap[ group[color][i] ] ])
				 isvisit[v] = true;

	int v = sum;
	if (sum % 2 == 0)
		sum /= 2;
	else
		sum = sum/2+1;

	for (int i = 0; i+sum <= v; i++)
		if (isvisit[sum+i])
			return sum+i;
}

int main()
{
	string s;

	while (cin >> m >> n && m||n)
	{
		cloth.clear();
		for (int i = 0; i < m; i++)
		{
			cin >> s;
			cloth[s] = i;
			group[i].clear();
		}

		for (int i = 1; i <= n; i++)
		{
			cin >> cap[i] >> s;
			group[ cloth[s] ].push_back(i);
		}

		int ans = 0;
		for (int i = 0; i < m; i++)
			ans += find(i);

		cout << ans << endl;
	}
	return 0;
}