题意：
给一个无向带权图，图上有不超过20个人和1个公园，现在这些人想到公园去集合，他们可以直接去公园也可以，和其他人一起坐车去公园（假设他们的车容量无限），但是这个公园停车场只有k个位置，现在要求他们到达公园所需要的总花费。
分析：
乍一看是最小生成树，但是停车场只有k个位置，所以就限定了公园节点只能最多连k个人，也就是说有一个点的度数是给定了的。
想了很久，第一感觉就是想其他生成树那样，肯定要把边删去，从环中选择更优解， 按套路来说。
但是想了很久不知道怎么处理。
不过，看到题目只有20个人，k<20，可以从n个人中选出k个人，与停车场连接，然后跑最小生成树。虽然很暴力，但最大复杂度才C(20,10)*mlogm，看起来不是很大啊，也的的确确A了，但是跑了接近2s，显然POJ数据真的很水。
其实这题是经典的问题，国家队论文里有讲，据说刘汝佳黑书里也有。推荐一个不错的博客：http://blog.csdn.net/nacl__/article/details/52052133
没错，我就是抄他代码的，自己写半天感觉非常搓，不过，他的代码没有删边，
这句话有问题，我修改了一下。
最近看了一下生成树和最短路的题，感觉天天发现新大陆啊。
代码：
标程
1 #include <set>
2 #include <map>
3 #include <list>
4 #include <cmath>
5 #include <queue>
6 #include <stack>
7 #include <vector>
8 #include <bitset>
9 #include <string>
10 #include <cctype>
11 #include <cstdio>
12 #include <cstring>
13 #include <cstdlib>
14 #include <iostream>
15 #include <algorithm>
16 // #include <unordered_map>
17
18 using namespace std;
19
20 typedef long long ll;
21 typedef unsigned long long ull;
22 typedef pair<int, int> pii;
23 typedef pair<ull, ull> puu;
24
25 #define inf (0x3f3f3f3f)
26 #define lnf (0x3f3f3f3f3f3f3f3f)
27 #define eps (1e-9)
28 #define fi first
29 #define se second
30
31 bool sgn(double a, string select, double b) {
32 if(select == "==")return fabs(a - b) < eps;
33 if(select == "!=")return fabs(a - b) > eps;
34 if(select == "<")return a - b < -eps;
35 if(select == "<=")return a - b < eps;
36 if(select == ">")return a - b > eps;
37 if(select == ">=")return a - b > -eps;
38 }
39
40
41 //--------------------------
42
43 const ll mod = 1000000007;
44 const int maxn = 30;
45
46 struct Edge {
47 int u, v, d;
48 Edge() {}
49 Edge(int a, int b, int c): u(a), v(b), d(c) {}
50 bool operator<(const Edge &e)const {
51 return d < e.d;
52 }
53 };
54
55 int n, m, k;
56 int cnt;
57 int ans;
58 int parent[maxn];
59 map<string, int> nodes;
60 vector<Edge>edges;
61 int g[maxn][maxn];
62 bool tree[maxn][maxn];
63 int minEdge[maxn];
64 Edge dp[maxn];
65
66 int find(int p) {
67 if(p == parent[p])return p;
68 else return parent[p] = find(parent[p]);
69 }
70
71 void un(int p, int q) {
72 parent[find(p)] = find(q);
73 }
74
75 void Kruskal() {
76 sort(edges.begin(), edges.end());
77 for(int i = 0; i < edges.size(); i++) {
78 int p = edges[i].u;
79 int q = edges[i].v;
80 if(p == 1 || q == 1)continue;
81 if(find(p) != find(q)) {
82 un(p, q);
83 tree[p][q] = tree[q][p] = 1;
84 ans += edges[i].d;
85 }
86 }
87 }
88
89 void dfs(int cur, int pre) {
90 for(int i = 2; i <= cnt; i++) {
91 if(i == pre || !tree[cur][i])continue;
92 if(dp[i].d == -1) {
93 if(dp[cur].d > g[cur][i])dp[i] = dp[cur];
94 else {
95 dp[i].u = cur;
96 dp[i].v = i;
97 dp[i].d = g[cur][i];
98 }
99 }
100 dfs(i, cur);
101 }
102 }
103
104
105 void init() {
106 memset(g, inf, sizeof(g));
107 memset(tree, 0, sizeof(tree));
108 memset(minEdge, inf, sizeof(minEdge));
109 m = 0;
110 cnt = 1;
111 ans = 0;
112 nodes["Park"] = 1;
113 for(int i = 0; i < maxn; i++) {
114 parent[i] = i;
115 }
116 }
117
118 void solve() {
119 scanf("%d", &n);
120 string s1, s2;
121 int d;
122 init();
123 for(int i = 1; i <= n; i++) {
124 cin >> s1 >> s2 >> d;
125 if(!nodes[s1])nodes[s1] = ++cnt;
126 if(!nodes[s2])nodes[s2] = ++cnt;
127 int u = nodes[s1];
128 int v = nodes[s2];
129 edges.push_back(Edge(u, v, d));
130 g[u][v] = g[v][u] = min(g[u][v], d);
131 }
132 scanf("%d", &k);
133 Kruskal();
134 int keyPoint[maxn];
135 for(int i = 2; i <= cnt; i++) {
136 if(g[1][i] != inf) {
137 int color = find(i);
138 if(minEdge[color] > g[1][i]) {
139 minEdge[color] = g[1][i];
140 keyPoint[color] = i;
141
142 }
143 }
144 }
145 for(int i = 1; i <= cnt; i++) {
146 if(minEdge[i] != inf) {
147 m++;
148 tree[1][keyPoint[i]] = tree[keyPoint[i]][1] = 1;
149 ans += g[1][keyPoint[i]];
150 }
151 }
152 for(int i = m + 1; i <= k; i++) {
153 memset(dp, -1, sizeof(dp));
154 dp[1].d = -inf;
155 for(int j = 2; j <= cnt; j++)
156 if(tree[1][j])
157 dp[j].d = -inf;
158 dfs(1, -1);
159 int idx, minnum = inf;
160 for(int j = 2; j <= cnt; j++) {
161 if(minnum > g[1][j] - dp[j].d) {
162 minnum = g[1][j] - dp[j].d;
163 idx = j;
164 }
165 }
166 if(minnum >= 0)
167 break;
168 tree[1][idx] = tree[idx][1] = 1;
169 tree[dp[idx].u][dp[idx].v] = tree[dp[idx].v][dp[idx].u] = 0;
170 ans += minnum;
171 }
172 printf("Total miles driven: %d\n", ans);
173 }
174
175 int main() {
176
177 #ifndef ONLINE_JUDGE
178 freopen("1.in", "r", stdin);
179 // freopen("1.out", "w", stdout);
180 #endif
181 // iostream::sync_with_stdio(false);
182 solve();
183 return 0;
184 }
暴力
1 #include <set>
2 #include <map>
3 #include <list>
4 #include <cmath>
5 #include <queue>
6 #include <stack>
7 #include <vector>
8 #include <bitset>
9 #include <string>
10 #include <cctype>
11 #include <cstdio>
12 #include <cstring>
13 #include <cstdlib>
14 #include <iostream>
15 #include <algorithm>
16 // #include <unordered_map>
17
18 using namespace std;
19
20 typedef long long ll;
21 typedef unsigned long long ull;
22 typedef pair<int, int> pii;
23 typedef pair<ull, ull> puu;
24
25 #define inf (0x3f3f3f3f)
26 #define lnf (0x3f3f3f3f3f3f3f3f)
27 #define eps (1e-9)
28 #define fi first
29 #define se second
30
31 bool sgn(double a, string select, double b) {
32 if(select == "==")return fabs(a - b) < eps;
33 if(select == "!=")return fabs(a - b) > eps;
34 if(select == "<")return a - b < -eps;
35 if(select == "<=")return a - b < eps;
36 if(select == ">")return a - b > eps;
37 if(select == ">=")return a - b > -eps;
38 }
39
40
41 //--------------------------
42
43 const ll mod = 1000000007;
44 const int maxn = 100010;
45
46 int n, k;
47
48 map<string, int> name;
49 int cnt;
50 int edn;
51 int pkn;
52
53 struct Edge {
54 int u, v, c;
55 };
56
57 bool cmp(Edge a, Edge b ) {
58 return a.c < b.c;
59 }
60
61
62 Edge raw_edges[500];
63 Edge park_edges[500];
64 Edge edges[500];
65
66 string a, b;
67 int x;
68
69 int ans = inf;
70
71 int max_park;
72
73 int par[500];
74 int findx(int x) {
75 if(par[x] == x)return x;
76 else return par[x] = findx(par[x]);
77 }
78
79
80 int Kruskal(int n) {
81 int m = edn + max_park;
82 for(int i = 0; i < n; i++) {
83 par[i] = i;
84 }
85 sort(edges, edges + m , cmp);
86 int cnt = 0;
87 int ans = 0;
88 for(int i = 0; i < m; i++) {
89 int u = edges[i].u;
90 int v = edges[i].v;
91 int c = edges[i].c;
92 int t1 = findx(u);
93 int t2 = findx(v);
94 if(t1 != t2) {
95 ans += c;
96 par[t1] = t2;
97 cnt++;
98 }
99 if(cnt == n - 1)break;
100 }
101 if(cnt < n - 1)return -1;
102 else return ans;
103 }
104
105
106 int select[30];
107
108 void dfs(int now, int pre) {
109 if(now >= max_park) {
110 for(int i = 0; i < edn; i++) {
111 edges[i] = raw_edges[i];
112 }
113 int con = edn;
114 for(int i = 0; i < now; i++) {
115 edges[con++] = park_edges[select[i]];
116 }
117 int mins = Kruskal(cnt);
118 // for(int i = edn; i < con; i++) {
119 // printf("%d %d\n", edges[i].u, edges[i].v );
120 // }
121 // printf("res = %d\n", mins );
122 if(mins != -1) {
123 ans = min(ans, mins);
124 }
125 return ;
126 }
127 for(int i = pre + 1; i < pkn; i++) {
128 select[now] = i;
129 dfs(now + 1, i);
130 }
131
132 }
133
134
135 void solve() {
136 while(cin >> n) {
137 cnt = edn = pkn = 0;
138 ans = inf;
139 name.clear();
140 for(int i = 0; i < n; i++) {
141 cin >> a >> b >> x;
142 if(name.find(a) == name.end()) {
143 name[a] = cnt++;
144 }
145 if(name.find(b) == name.end()) {
146 name[b] = cnt++;
147 }
148 if(a == "Park" || b == "Park") {
149 park_edges[pkn++] = Edge{name[a], name[b], x};
150 } else {
151 raw_edges[edn++] = Edge{name[a], name[b], x};
152 }
153 }
154 scanf("%d", &k);
155 max_park = min(pkn, k);
156 dfs(0, -1);
157 printf("Total miles driven: %d\n", ans);
158 }
159
160
161 }
162
163 int main() {
164
165 #ifndef ONLINE_JUDGE
166 freopen("1.in", "r", stdin);
167 freopen("1.out", "w", stdout);
168 #endif
169 // iostream::sync_with_stdio(false);
170 solve();
171 return 0;
172 }
时间： 2024-12-20 00:09:04