HDU 5195 DZY Loves Topological Sorting 拓扑排序

题目链接：

hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5195

bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=573&pid=1002

题解：

1、拓扑排序+贪心

 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<vector>
 #include<queue>
 using namespace std;

 const int maxn = 1e5 + ;
 const int INF = 0x3f3f3f3f;

 int n, m, k;

 struct Node {
     int v, flag;
     Node(int v,int flag=):v(v),flag(flag){}
     Node() { flag = ; }
 };

 vector<Node> head[maxn];
 int ind[maxn],done[maxn];

 void init() {
     for (int i = ; i < n; i++) {
         ind[i] = done[i]=;
         head[i].clear();
     }
 }

 int main() {
     while (scanf("%d%d%d", &n, &m, &k) ==  && n) {
         init();
         for (int i = ; i < m; i++) {
             int u, v;
             scanf("%d%d", &u, &v); u--, v--;
             ind[v]++;
             head[u].push_back(Node(v,));
         }

         priority_queue<int> pq;

         //删边
         for (int i = n - ; i >= ; i--) {
             if (k >= ind[i]) {
                 //将i的父亲ui中，满足ui<i，即边（ui,i）删了，这里要注意，对于边（ui,i),ui>i的边，根本不用删
                 k -= ind[i];
                 pq.push(i);
                 for (int j = ; j < head[i].size(); j++) {
                     Node &nd = head[i][j];
                     if (i > nd.v) {
                         //把边（i,v）删了，拓扑排序的时候不能再走这条边了
                         nd.flag = ;
                         ind[nd.v]--;
                     }
                 }
             }
         }

         vector<int> ans;
         //拓扑排序
         while (!pq.empty()) {
             int u = pq.top(); pq.pop();
             if (done[u]) continue;
             ans.push_back(u);
             done[u] = ;
             for (int i = ; i < head[u].size(); i++) {
                 Node& nd = head[u][i];
                 if (done[nd.v]||nd.flag) continue;
                 ind[nd.v]--;
                 if (ind[nd.v] == ) pq.push(nd.v);
             }
         }

         printf("%d", ans[]+);
         for (int i = ; i < ans.size(); i++) printf(" %d", ans[i]+);
         printf("\n");
     }
     return ;
 }
 /*
 5 3 1
 4 3
 1 3
 3 2

 5 3 0
 4 3
 1 3
 3 2
 */

2、线段树+贪心

对于节点i，入度为ind[i]，则可以这样贪心：对于1<=i<=n，求最大的i使得ind[i]<=k，我们可以把入度数组做成线段树，维护最小入度，二分查找，优先搜右边（右边的i会更大），找到以后，吧对于的子节点vj的入度减1，k-=ind[i]。这样一直做n次就能得到答案。

 #include<iostream>
 #include<cstdio>
 #include<vector>
 #include<algorithm>
 #define lson (o<<1)
 #define rson ((o<<1)+1)
 #define M (l+(r-l)/2)
 using namespace std;

 const int maxn = 1e5 + ;
 const int INF = 0x3f3f3f3f;

 int n, m, k;

 vector<int> head[maxn];
 int ind[maxn<<],minv[maxn<<];

 int _pos, _v;
 void update(int o, int l, int r) {
     if (l==r) {
         ind[o] += _v;
         minv[o] = ind[o];
     }
     else {
         if (_pos <= M) update(lson, l, M);
         else update(rson, M + , r);
         minv[o] = min(minv[lson], minv[rson]);
     }
 }

 void query(int o, int l, int r,int &res) {
     if (l == r) {
         if (ind[o] <= k) {
             k -= ind[o];
             res = l;
         }
     }
     else {
         //printf("lson:%d,rson:%d\n",minv[lson],minv[rson]);
         if (k >= minv[rson]) query(rson, M + , r,res);
         else if (k >= minv[lson]) query(lson, l, M,res);
     }
 }

 void init() {
     for (int i = ; i <= n; i++) head[i].clear();
     memset(ind, , sizeof(ind));
     memset(minv,,sizeof(minv));
 }

 int main() {
     while (scanf("%d%d%d", &n, &m, &k) ==  && n) {
         init();
         for (int i = ; i < m; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             head[u].push_back(v);
             _pos = v, _v = ;
             update(, , n);
         }
         //puts("after update");
         vector<int> ans;
         for (int i = ; i < n; i++) {
             int res;
             query(, , n, res);
             //puts("after first qurey!");
             //printf("res:%d\n", res);
             ans.push_back(res);
             for (int j = ; j < head[res].size(); j++) {
                 _pos = head[res][j], _v = -;
                 update(, , n);
                 //puts("after first upate");
             }
             _pos = res, _v = INF;
             update(, , n);
         }
         //puts("ans is zero");
         printf("%d", ans[]);
         for (int i = ; i < ans.size(); i++) printf(" %d", ans[i]);
         printf("\n");
     }
     return ;
 }