洛谷P1041 传染病控制

解：搜索......

我们可以每次选择分支少的搜索，或者说，贪心的搜索当前更优的决策。

每一层把能剪的点搞出来，按照度数/SIZ排序，然后依次搜索。加个最优化剪枝就完事了。

 1 #include <bits/stdc++.h>
 2 
 3 const int N = 310, INF = 0x3f3f3f3f;
 4 
 5 struct Edge {
 6     int nex, v;
 7 }edge[N << 1]; int tp;
 8 
 9 int e[N], siz[N], in[N], n, ans = INF;
10 int v[N][N];
11 bool del[N], vis[N];
12 
13 inline bool cmp(const int &a, const int &b) {
14     //return in[a] > in[b];
15     return siz[a] > siz[b];
16 }
17 
18 inline void add(int x, int y) {
19     tp++;
20     edge[tp].v = y;
21     edge[tp].nex = e[x];
22     e[x] = tp;
23     return;
24 }
25 
26 void DFS_1(int x, int f) {
27     siz[x] = 1;
28     for(int i = e[x]; i; i = edge[i].nex) {
29         int y = edge[i].v;
30         if(y == f) continue;
31         DFS_1(y, x);
32         siz[x] += siz[y];
33     }
34     return;    
35 }
36 
37 void DFS(int k, int d) {
38     if(k >= ans) return;
39     v[d][0] = 0;
40     for(int x = 1; x <= n; x++) {
41         if(del[x] || !vis[x]) continue;
42         /// vis[x] 
43         for(int i = e[x]; i; i = edge[i].nex) {
44             int y = edge[i].v;
45             if(vis[y] || del[y]) continue;
46             v[d][++v[d][0]] = y;
47         }
48     }
49     
50     if(!v[d][0]) {
51         ans = std::min(ans, k);
52         return;
53     }
54     
55     std::sort(v[d] + 1, v[d] + v[d][0] + 1, cmp);
56     for(int i = 1; i <= v[d][0]; i++) {
57         vis[v[d][i]] = 1;
58     }
59     k += v[d][0] - 1;
60     for(int i = 1; i <= v[d][0]; i++) {
61         int x = v[d][i];
62         del[x] = 1;
63         vis[x] = 0;
64         DFS(k, d + 1);
65         vis[x] = 1;
66         del[x] = 0;
67     }
68     for(int i = 1; i <= v[d][0]; i++) {
69         vis[v[d][i]] = 0;
70     }
71     return;
72 }
73 
74 int main() {
75     
76 //    freopen("disease.in", "r", stdin);
77 //    freopen("disease.out", "w", stdout);
78 
79     int m;
80     scanf("%d%d", &n, &m);
81     for(int i = 1, x, y; i <= m; i++) {
82         scanf("%d%d", &x, &y);
83         add(x, y); add(y, x);
84         in[x]++; in[y]++;
85     }
86     DFS_1(1, 0);
87     vis[1] = 1;
88     DFS(1, 0);
89     
90     printf("%d\n", ans);
91     return 0;
92 }

AC代码