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项目时间安排的解决方案

项目时间安排问题可以通过拓扑排序来解决。在传统的贪心算法中，项目之间没有明确的依赖关系，这种方法并不适用。但当项目间存在明确的依赖关系时，我们可以将其建模为有向图，通过拓扑排序来确定项目的执行顺序。

拓扑排序的基本思想是：

以下是基于C语言的拓扑排序实现代码：

int* topSort(int* returnSize, int* deg, int** graph, int* graphColSize, int* items, int itemsSize) {    *returnSize = 0;    int Q[itemsSize];    int left = 0, right = 0;    for (int i = 0; i < itemsSize; i++) {        if (deg[items[i]] == 0) {            Q[right++] = items[i];        }    }    int* res = malloc(sizeof(int) * itemsSize);    while (left < right) {        int u = Q[left++];        res[(*returnSize)++] = u;        for (int i = 0; i < graphColSize[u]; i++) {            int v = graph[u][i];            if (--deg[v] == 0) {                Q[right++] = v;            }        }    }    if (*returnSize == itemsSize) {        return res;    }    *returnSize = 0;    return NULL;}

拓扑排序的核心逻辑包括：

在实际项目中，我们还需要处理组间依赖关系。这可以通过将项目分为不同的组，并为每个组生成一个独立的拓扑排序。具体实现步骤如下：

以下是C++语言的实现代码：

class Solution {    vector
   
     sortItems(int n, int m, vector
    
     & group, const vector
     
      
       >& beforeItems) {        vector
       
         groupItem(n + m);        vector
        
          groupItemColSize(n + m); vector
         
           groupItemColCapacity(n + m); for (int i = 0; i < n + m; i++) { groupItem[i] = i; groupItemColSize[i] = 0; groupItemColCapacity[i] = 0; } vector
          
            groupGraph(n + m); vector
           
             groupGraphColSize(n + m); vector
            
              groupGraphColCapacity(n + m); for (int i = 0; i < n + m; i++) { groupGraph[i] = new int; groupGraphColSize[i] = 0; groupGraphColCapacity[i] = 0; } vector
             
               itemGraph(n); vector
              
                itemGraphColSize(n); vector
               
                 itemGraphColCapacity(n); for (int i = 0; i < n; i++) { itemGraph[i] = new int; itemGraphColSize[i] = 0; itemGraphColCapacity[i] = 0; } vector
                
                  groupDegree(n + m, 0); vector
                 
                   itemDegree(n, 0); vector
                  
                    id(n + m); for (int i = 0; i < n + m; i++) { id[i] = i; } int leftId = m; for (int i = 0; i < n; i++) { if (group[i] == -1) { group[i] = leftId++; } if (groupItemColSize[group[i]] == groupItemColCapacity[group[i]]) { int* x = &groupItemColCapacity[group[i]]; (*x) = (*x) ? (*x) * 2 : 1; groupItem[group[i]] = new int(*groupItemColCapacity[group[i]]); } groupItem[group[i]]->push_back(i); } for (int i = 0; i < n; i++) { int curGroupId = group[i]; for (int j = 0; j < beforeItems[i].size(); j++) { int item = beforeItems[i][j]; int beforeGroupId = group[item]; if (beforeGroupId == curGroupId) { itemDegree[i]++; if (itemGraphColSize[item] == itemGraphColCapacity[item]) { int* x = &itemGraphColCapacity[item]; (*x) = (*x) ? (*x) * 2 : 1; itemGraph[item] = new int(*itemGraphColCapacity[item]); } itemGraph[item]->push_back(i); } else { groupDegree[curGroupId]++; if (groupGraphColSize[beforeGroupId] == groupGraphColCapacity[beforeGroupId]) { int* x = &groupGraphColCapacity[beforeGroupId]; (*x) = (*x) ? (*x) * 2 : 1; groupGraph[beforeGroupId] = new int(*groupGraphColCapacity[beforeGroupId]); } groupGraph[beforeGroupId]->push_back(curGroupId); } } } queue
                   
                     q; for (int i = 0; i < n; i++) { if (itemDegree[i] == 0) { q.push(i); } } int countItem = 0; vector
                    
                      g_items; while (!q.empty()) { int u = q.front(); q.pop(); countItem++; g_items.push_back(u); for (int v = 0; v < itemGraphColSize[u]; v++) { int item = itemGraph[u][v]; if (--itemDegree[item] == 0) { q.push(item); } } } if (countItem != n) { return vector
                     
                      (); } queue
                      
                        groupQ; for (int i = 0; i < m; i++) { if (groupDegree[i] == 0) { groupQ.push(i); } } int countGroup = 0; vector
                       
                         g_order; while (!groupQ.empty()) { int g = groupQ.front(); groupQ.pop(); countGroup++; g_order.push_back(g); for (int h = 0; h < groupGraphColSize[g]; h++) { int group = groupGraph[g][h]; if (--groupDegree[group] == 0) { groupQ.push(group); } } } if (countGroup != m) { return vector
                        
                         (); } vector
                         
                           ans; int idx = 0; for (int g : g_order) { for (int item : *groupItem[g]) { ans[idx++] = item; } } return ans; }}
                         
                        
                       
                      
                     
                    
                   
                  
                 
                
               
              
             
            
           
          
         
        
       
      
     
    
   

以下是Python实现：

import collectionsclass Solution:    def sortItems(self, n, m, group, beforeItems):        group2Items = collections.defaultdict(set)        index = m        for i in range(n):            if group[i] == -1:                group[i] = index                group2Items[index].add(i)                index += 1            else:                group2Items[group[i]].add(i)        beforeGroupsDic = {}        preGroupsDic = {}        for i in range(index):            beforeGroupsDic[i] = set()            preGroupsDic[i] = set()        for i in range(n):            for j in beforeItems[i]:                beforeGroupsDic[group[i]].add(group[j])                preGroupsDic[group[j]].add(group[i])        for i in range(index):            if i in beforeGroupsDic[i]:                beforeGroupsDic[i].remove(i)                preGroupsDic[i].remove(i)        res = []        while beforeGroupsDic:            groups = [key for key, values in beforeGroupsDic.items() if not beforeGroupsDic[key]]            if not groups:                return []            for g in groups:                beforeGroupsDic.pop(g)                for g1 in preGroupsDic[g]:                    beforeGroupsDic[g1].remove(g)                preGroupsDic.pop(g)            for key in groups:                its = list(group2Items[key])                beforeItemsDic = {}                preItemsDic = {}                for i in its:                    beforeItemsDic[i] = set()                    preItemsDic[i] = set()                for i in its:                    for j in beforeItems[i]:                        if group[i] == group[j]:                            beforeItemsDic[i].add(j)                            preItemsDic[j].add(i)                while beforeItemsDic:                    items = [k for k, v in beforeItemsDic.items() if not beforeItemsDic[k]]                    if not items:                        return []                    res += items                    for item in items:                        beforeItemsDic.pop(item)                        for item1 in preItemsDic[item]:                            if item1 in beforeItemsDic:                                beforeItemsDic[item1].remove(item)                    preItemsDic.pop(item)        return res

以上代码实现了项目时间安排的解决方案。通过拓扑排序，我们可以有效地确定项目的执行顺序，确保项目按时完成。

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