poj 3422 Kaka's Matrix Travels 费用流
给一个n*n的矩阵， 从左上角出发， 走到右下角， 然后在返回左上角，这样算两次。 一共重复k次， 每个格子有值， 问能够取得的最大值是多少， 一个格子的值只能取一次， 取完后变为0。
费用流第一题， 将每个格子拆为两个点， u向u'连一条容量为1， 费用为格子的值的边， u向u'再连一条容量为k-1， 费用为0的边。u'向他右边和下边的格子连一条容量为k， 费用为0的边， 跑一遍费用流就可以。
1 #include <iostream> 2 #include <vector> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <cmath> 7 #include <map> 8 #include <set> 9 #include <string> 10 #include <queue> 11 using namespace std; 12 #define pb(x) push_back(x) 13 #define ll long long 14 #define mk(x, y) make_pair(x, y) 15 #define lson l, m, rt<<1 16 #define mem(a) memset(a, 0, sizeof(a)) 17 #define rson m+1, r, rt<<1|1 18 #define mem1(a) memset(a, -1, sizeof(a)) 19 #define mem2(a) memset(a, 0x3f, sizeof(a)) 20 #define rep(i, a, n) for(int i = a; i<n; i++) 21 #define ull unsigned long long 22 typedef pair<int, int> pll; 23 const double PI = acos(-1.0); 24 const double eps = 1e-8; 25 const int mod = 1e9+7; 26 const int inf = 1061109567; 27 const int dir[][2] = { {1, 0}, {0, 1}, {0, -1}, {0, 1} }; 28 const int maxn = 2e5+5; 29 int num, head[maxn*2], s, t, n, k, nn, dis[maxn], flow, cost, cnt, cap[maxn], q[maxn], cur[maxn], vis[maxn]; 30 struct node 31 { 32 int to, nextt, c, w; 33 node(){} 34 node(int to, int nextt, int c, int w):to(to), nextt(nextt), c(c), w(w) {} 35 }e[maxn*2]; 36 int spfa() { 37 int st, ed; 38 st = ed = 0; 39 mem2(dis); 40 ++cnt; 41 dis[s] = 0; 42 cap[s] = inf; 43 cur[s] = -1; 44 q[ed++] = s; 45 while(st<ed) { 46 int u = q[st++]; 47 vis[u] = cnt-1; 48 for(int i = head[u]; ~i; i = e[i].nextt) { 49 int v = e[i].to, c = e[i].c, w = e[i].w; 50 if(c && dis[v]>dis[u]+w) { 51 dis[v] = dis[u]+w; 52 cap[v] = min(c, cap[u]); 53 cur[v] = i; 54 if(vis[v] != cnt) { 55 vis[v] = cnt; 56 q[ed++] = v; 57 } 58 } 59 } 60 } 61 if(dis[t] == inf) 62 return 0; 63 cost += dis[t]*cap[t]; 64 flow += cap[t]; 65 for(int i = cur[t]; ~i; i = cur[e[i^1].to]) { 66 e[i].c -= cap[t]; 67 e[i^1].c += cap[t]; 68 } 69 return 1; 70 } 71 int mcmf() { 72 flow = cost = 0; 73 while(spfa()) 74 ; 75 return cost; 76 } 77 void add(int u, int v, int c, int val) { 78 e[num] = node(v, head[u], c, -val); head[u] = num++; 79 e[num] = node(u, head[v], 0, val); head[v] = num++; 80 } 81 void input() { 82 int x; 83 for(int i = 0; i<n; i++) { 84 for(int j = 0; j<n; j++) { 85 scanf("%d", &x); 86 add(i*n+j, i*n+j+nn, 1, x); 87 add(i*n+j, i*n+j+nn, k-1, 0); 88 } 89 } 90 add(s, 0, k, 0); 91 add(2*nn-1, t, k, 0); 92 for(int i = 0; i<n; i++) { 93 for(int j = 0; j<n; j++) { 94 for(int k1 = 0; k1<2; k1++) { 95 int x = dir[k1][0]+i; 96 int y = dir[k1][1]+j; 97 if(x>=0&&x<n&&y>=0&&y<n) { 98 add(i*n+j+nn, x*n+y, k, 0); 99 } 100 } 101 } 102 } 103 } 104 void init() { 105 mem1(head); 106 num = cnt = 0; 107 mem(vis); 108 } 109 int main() 110 { 111 while(~scanf("%d%d", &n, &k)) { 112 init(); 113 nn = n*n; 114 s = 2*nn, t = s+1; 115 input(); 116 int ans = mcmf(); 117 printf("%d\n", -ans); 118 } 119 return 0; 120 }
