题目：
http://acm.hdu.edu.cn/showproblem.php?pid=4862
题意：
给你一个n*m的矩阵，填充着0-9的数字，每次能从一个点出发，到它的右边或者下边的点，花费为|x1-x2|+|y1-y2|-1，如果跳跃的起点和终点的数字相同，则获得这个数字的收益，不能走已经走过的点
有K次重新选择起点的机会
如果可以走遍所有点，则输出最大的价值(收益-花费)
否则，输出-1
方法：
最小K路径覆盖，最小费用最大流
建图：
每个点拆为2点：X部和Y部，(a,b)表示流量a，费用b
源点与X部每个点连(1,0)的边
Y部每个点与汇点连(1,0)的边
X部的点如果可以到Y部的点，则连(1,花费-收益)的边
源点与一个新点连(k,0)的边，新点与Y部每个点连(1,0)的边
结果：
如果满流，则输出0-费用
否则，输出-1
代码：
1 // #pragma comment(linker, "/STACK:102400000,102400000")
2 #include <cstdio>
3 #include <iostream>
4 #include <cstring>
5 #include <string>
6 #include <cmath>
7 #include <set>
8 #include <list>
9 #include <map>
10 #include <iterator>
11 #include <cstdlib>
12 #include <vector>
13 #include <queue>
14 #include <stack>
15 #include <algorithm>
16 #include <functional>
17 using namespace std;
18 typedef long long LL;
19 #define ROUND(x) round(x)
20 #define FLOOR(x) floor(x)
21 #define CEIL(x) ceil(x)
22 // const int maxn = 210;
23 // const int maxm = 200010;
24 // const int inf = 0x3f3f3f3f;
25 const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
26 const double INF = 1e30;
27 const double eps = 1e-6;
28 const int P[4] = {0, 0, -1, 1};
29 const int Q[4] = {1, -1, 0, 0};
30 const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
31 const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
32
33 /**
34 *最小(大)费用最大流：SPFA增广路($O(w*O(SPFA))$)
35 *最大费用：费用取反addEdge(,,,-cost);
36 *输入：图(链式前向星),n(顶点个数,包含源汇),s(源),t(汇)
37 *输出：minCostMaxflow(int s, int t, int &cost)返回流量, cost为费用
38 *打印路径方法：按反向边(i&1)的flow 找，或者按边的flow找
39 */
40 const int maxn = 210;
41 const int maxm = 200010;
42 const int inf = 0x3f3f3f3f;
43 struct Edge
44 {
45 int u, v;
46 int cap, flow;
47 int cost;
48 int next;
49 } edge[maxm];
50 int head[maxn], en; //需初始化
51 int n, m;
52 int st, ed;
53 bool vis[maxn];
54 int pre[maxn], dis[maxn];
55 void addse(int u, int v, int cap, int flow, int cost)
56 {
57 edge[en].u = u;
58 edge[en].v = v;
59 edge[en].cap = cap;
60 edge[en].flow = flow;
61 edge[en].cost = cost;
62 edge[en].next = head[u];
63 head[u] = en++;
64 }
65 void adde(int u, int v, int cap, int cost)
66 {
67 addse(u, v, cap, 0, cost);
68 addse(v, u, 0, 0, -cost); //注意加反向0 边
69 }
70 bool spfa(int s, int t)
71 {
72 queue<int>q;
73 for (int i = 0; i < n; i++)
74 {
75 dis[i] = inf;
76 vis[i] = false;
77 pre[i] = -1;
78 }
79 dis[s] = 0;
80 vis[s] = true;
81 q.push(s);
82 while (!q.empty())
83 {
84 int u = q.front();
85 q.pop();
86 vis[u] = false;
87 for (int i = head[u]; i != -1; i = edge[i].next)
88 {
89 int v = edge[i].v;
90 if (edge[i].cap > edge[i].flow &&
91 dis[v] > dis[u] + edge[i].cost )
92 {
93 dis[v] = dis[u] + edge[i].cost;
94 pre[v] = i;
95 if (!vis[v])
96 {
97 vis[v] = true;
98 q.push(v);
99 }
100 }
101 }
102 }
103 if (pre[t] == -1)return false;
104 else return true;
105 }
106 int minCostMaxflow(int s, int t, int &cost)//返回流量, cost为费用
107 {
108 int flow = 0;
109 cost = 0;
110 while (spfa(s, t))
111 {
112 int Min = inf;
113 for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].v])
114 {
115 if (Min > edge[i].cap - edge[i].flow)
116 Min = edge[i].cap - edge[i].flow;
117 }
118 for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].v])
119 {
120 edge[i].flow += Min;
121 edge[i ^ 1].flow -= Min;
122 cost += edge[i].cost * Min;
123 }
124 flow += Min;
125 }
126 return flow;
127 }
128 int N, M, K;
129 int kase;
130 int mtx[maxn][maxn];
131 int disxy(int x1, int y1, int x2, int y2)
132 {
133 return abs(x1 - x2) + abs(y1 - y2) - 1;
134 }
135 void build()
136 {
137 n = 3 + N * M * 2;
138 st = 0, ed = 1;
139 for (int i = 0; i < N; i++)
140 {
141 for (int j = 0; j < M; j++)
142 {
143 adde(st, i * M + j + 3, 1, 0);
144 }
145 }
146 adde(st, 2, K, 0);
147 for (int i = 0; i < N; i++)
148 {
149 for (int j = 0; j < M; j++)
150 {
151 adde(2, i * M + j + N * M + 3, 1, 0);
152 adde(i * M + j + N * M + 3, ed, 1, 0);
153 }
154 }
155 for (int i = 0; i < N; i++)
156 {
157 for (int j = 0; j < M; j++)
158 {
159 for (int h = i + 1; h < N; h++)
160 {
161 if (mtx[i][j] == mtx[h][j])
162 {
163 adde(i * M + j + 3, h * M + j + N * M + 3, 1, h - i - 1 - mtx[i][j]);
164 // cout << i << " " << j << " " << h << " " << j << endl;
165 }
166 else
167 {
168 adde(i * M + j + 3, h * M + j + N * M + 3, 1, h - i - 1);
169 }
170 }
171 for (int h = j + 1; h < M; h++)
172 {
173 if (mtx[i][j] == mtx[i][h])
174 {
175 adde(i * M + j + 3, i * M + h + N * M + 3, 1, h - j - 1 - mtx[i][j]);
176 // cout << i << " " << j << " " << i << " " << h << endl;
177 }
178 else
179 {
180 adde(i * M + j + 3, i * M + h + N * M + 3, 1, h - j - 1);
181 }
182 }
183 }
184 }
185 }
186 void init()
187 {
188 memset(head, -1, sizeof(head));
189 en = 0;
190 kase++;
191 }
192 void input()
193 {
194 scanf("%d%d%d", &N, &M, &K);
195 for (int i = 0; i < N; i++)
196 {
197 char str[maxn];
198 scanf("%s", str);
199 for (int j = 0; j < M; j++)
200 {
201 mtx[i][j] = str[j] - ‘0‘;
202 }
203 }
204 }
205 void debug()
206 {
207 //
208 }
209 void solve()
210 {
211 build();
212 int cost;
213 int flow = minCostMaxflow(st, ed, cost);
214 // cout << "flow,cost: " << flow << " " << cost << endl;
215 if (flow == N * M)
216 {
217 printf("Case %d : %d\n", kase, -cost);
218 }
219 else
220 {
221 printf("Case %d : %d\n", kase, -1);
222 }
223 }
224 void output()
225 {
226 //
227 }
228 int main()
229 {
230 // int size = 256 << 20; // 256MB
231 // char *p = (char *)malloc(size) + size;
232 // __asm__("movl %0, %%esp\n" :: "r"(p));
233
234 // std::ios_base::sync_with_stdio(false);
235 #ifndef ONLINE_JUDGE
236 freopen("in.cpp", "r", stdin);
237 #endif
238
239 kase = 0;
240 int T;
241 scanf("%d", &T);
242 while (T--)
243 {
244 init();
245 input();
246 solve();
247 output();
248 }
249 return 0;
250 }
HDU 4862
HDU 4862 Jump (2014-多校1-1002，最小K路径覆盖，最小费用最大流)
时间： 2024-10-19 02:06:00