noip2008 传纸条 费用流版
最近学了下费用流，然后又从网上听说了传纸条可以用费用流完成，于是就试了一下。结果调了半天终于通过了。其原理很简单，就是在图中求两条不相交路径，使途经点权和最大。方法就是把原图中的每个点x拆成x1和x2，对于每个点x和它右面或者下面的点y，建立起止点为x2,y1,费用为0,流量为1的边。并且建立起止点为x1,x2,费用为点权，流量为1的边。特别地，起点（1,1）和终点（n,m）点内的边流量应为2.然后求图的最大费用最大流即可。
一直在费用流增广那里卡了很久……总是报SISEGV，后来发现条件应该是i!=S而不是i!=0，然后就A了。这种方法比DP高效，但是代码长度和思维难度…………所以这种方法对于NOIP来说，简直是高射炮打蚊子。
View Code
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6
7 const int inf = 0x3f3f3f3f;
8 const int maxn = 5010;
9 const int maxm = 20010;
10 struct Node {
11 int end, val, cost;
12 Node *next, *op;
13 }edge[maxm], *head[maxn], *p, *fe[maxn];
14 int S, T, n, m, ans;
15 int dist[maxn], que[maxn], ft[maxn];
16 bool inq[maxn];
17
18 inline int bh(int i, int j, int flag) {
19 return ((i - 1) * m + j) * 2 + flag - 1;
20 }
21 inline void add_edge(int a, int b, int v, int co) {
22 p ++;
23 p -> end = b;
24 p -> val = v;
25 p -> cost = co;
26 p -> next = head[a];
27 head[a] = p;
28 p ++;
29 p -> end = a;
30 p -> val = 0;
31 p -> cost = -co;
32 p -> next = head[b];
33 head[b] = p;
34 head[a] -> op = head[b];
35 head[b] -> op = head[a];
36 }
37 inline void init() {
38 p = edge; int fy;
39 scanf("%d%d", &n, &m);
40 S = 1, T = n * m * 2;
41 for (int i = 1; i <= n; i ++)
42 for (int j = 1; j <= m; j ++) {
43 scanf("%d", &fy);
44 if (i == 1 && j == 1) {
45 add_edge(bh(1, 1, 0), bh(1, 1, 1), 2, 0);
46 } else if (i == n && j == m){
47 add_edge(bh(n, m, 0), bh(n, m, 1), 2, 0);
48 } else {
49 add_edge(bh(i, j, 0), bh(i, j, 1), 1, fy);
50 }
51 if (j < m) add_edge(bh(i, j, 1), bh(i, j + 1, 0), 1, 0);
52 if (i < n) add_edge(bh(i, j, 1), bh(i + 1, j, 0), 1, 0);
53 }
54 }
55 inline bool spfa() {
56 for (int i = S; i <= T; i ++)
57 dist[i] = -inf;
58 dist[S] = 0;
59 int open = 0, close = 0;
60 que[++ open] = S; inq[S] = true;
61 while (open != close) {
62 close ++; close %= maxn;
63 int t = que[close]; inq[t] = false;
64 for (Node *p = head[t]; p; p = p -> next) {
65 int end = p -> end;
66 if (p -> val && dist[t] + p -> cost > dist[end]) {
67 dist[end] = dist[t] + p -> cost;
68 ft[end] = t;
69 fe[end] = p;
70 if (!inq[end]) {
71 open ++; open %= maxn;
72 que[open] = end; inq[end] = true;
73 }
74 }
75 }
76 }
77 return dist[T] != -inf;
78 }
79 inline void augment() {
80 int min_f = inf;
81 for (int i = T; i != S; i = ft[i])
82 min_f = min(min_f, fe[i] -> val);
83 for (int i = T; i != S; i = ft[i]) {
84 fe[i] -> val -= min_f;
85 fe[i] -> op -> val += min_f;
86 ans += min_f * fe[i] -> cost;
87 }
88 }
89 inline int work() {
90 ans = 0;
91 while (spfa())
92 augment();
93 return ans;
94 }
95 int main() {
96 init();
97 printf("%d\n", work());
98 return 0;
99 }
