POJ 3211 Washing Clothes (平衡划分&01背包)

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http://poj.org/problem?id=3211

既然两个人可以同时洗同色的衣服，那么对每种颜色进行分析，转化为在sumtime/2的时间内能洗至多多久的衣服，则洗这种颜色的衣服至少用时sumtime-01pack(sumtime/2)

完整代码：

/*32ms,968KB*/#include<cstdio>#include<cstring>#include<map>#include<string>using namespace std;int cost[15][105], num[15], dp[100005];int _01pack(int V, int n, int kind){memset(dp, 0, sizeof(dp));int i, j;for (i = 0; i < n; i++)for (j = V; j >= cost[kind][i]; --j)dp[j] = max(dp[j], dp[j - cost[kind][i]] + cost[kind][i]);return dp[V];}int main(){int m, n, i, j, time, id, total_t, ans;char color[15];map<string, int> ma;while (scanf("%d%d", &m, &n), m){ma.clear();memset(num, 0, sizeof(num));memset(cost, 0, sizeof(cost));for (i = 0; i < m; ++i){scanf("%s", color);ma[color] = i;}for (i = 0; i < n; ++i){scanf("%d%s", &time, color);id = ma[color];cost[id][num[id]++] = time;}ans = 0;for (i = 0; i < m; ++i){total_t = 0;for (j = 0; j < num[i]; ++j)total_t += cost[i][j];ans += total_t - _01pack(total_t >> 1, num[i], i);}printf("%d\n", ans);}return 0;}

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