poj3422 Kaka's Matrix Travels

传送门

这个题一开始感觉像两条路径和 毕竟求的东西名字都一样

后来发现不行 因为那个是强制限制一条边只能走一次

这个是以后还可以走但是没有价值了

所以考虑一下拆点

不同的地方在于每个点只能走一次有权值的路径

所以拆的点之间建两条边 一条有权值流量为1 一条没权值流量为k-1

然后起点连源 终点连汇 左上右部图连右下左部图都是套路

注意求最大费用 取相反数即可

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<queue>
 6 #define ms(a,b) memset(a,b,sizeof a)
 7 #define rep(i,a,n) for(int i = a;i <= n;i++)
 8 #define per(i,n,a) for(int i = n;i >= a;i--)
 9 #define inf 2147483647
10 using namespace std;
11 typedef long long ll;
12 typedef double D;
13 #define eps 1e-8
14 ll read() {
15     ll as = 0,fu = 1;
16     char c = getchar();
17     while(c < '0' || c > '9') {
18         if(c == '-') fu = -1;
19         c = getchar();
20     }
21     while(c >= '0' && c <= '9') {
22         as = as * 10 + c - '0';
23         c = getchar();
24     }
25     return as * fu;
26 }
27 //head
28 const int N = 10005;
29 const int M = 100005;
30 int s = N-1,t = N-2;
31 int head[N],nxt[M],mo[M],cnt = 1;
32 int cst[M],flw[M];
33 void _add(int x,int y,int w,int f) {
34     mo[++cnt] = y;
35     nxt[cnt] = head[x];
36     head[x] = cnt;
37     cst[cnt] = w,flw[cnt] = f;
38 }
39 void add(int x,int y,int f,int w) {
40     if(x^y)_add(x,y,w,f),_add(y,x,-w,0);
41 }
42 
43 int dis[N],flow[N];
44 int pre[N],lst[N];
45 bool vis[N];
46 bool spfa() {
47     ms(dis,70),ms(flow,70),ms(vis,0);
48     queue<int> q;
49     q.push(s),pre[t] = dis[s] = 0,vis[s] = 1;
50     while(!q.empty()) {
51         int x = q.front();
52         q.pop(),vis[x] = 0;
53         for(int i = head[x];i;i = nxt[i]) {
54             int sn = mo[i];
55             if(dis[sn] > dis[x] + cst[i] && flw[i]) {
56                 dis[sn] = dis[x] + cst[i];
57                 pre[sn] = x,lst[sn] = i;
58                 flow[sn] = min(flow[x],flw[i]);
59                 if(!vis[sn]) vis[sn] = 1,q.push(sn);
60             }
61         }
62     }
63     return pre[t];
64 }
65 
66 int maxx,minn;
67 void MCMF() {
68     maxx = minn = 0;
69     while(spfa()) {
70         maxx += flow[t],minn += dis[t] * flow[t];
71         int x = t;
72         while(x ^ s) {
73             flw[lst[x]] -= flow[t];
74             flw[lst[x] ^ 1] += flow[t];
75             x = pre[x];
76         }
77     }
78 }
79 
80 int n,k;
81 int idx(int x,int y,int f) {return (x-1) * n + y + f * n * n;}
82 
83 void build() {
84     n = read(),k = read();
85     rep(i,1,n) rep(j,1,n) add(idx(i,j,0),idx(i,j,1),1,-read());
86     rep(i,1,n) rep(j,1,n) add(idx(i,j,0),idx(i,j,1),k-1,0);
87     add(s,idx(1,1,0),k,0);
88     add(idx(n,n,1),t,k,0);
89     rep(i,1,n) rep(j,1,n) {
90         if(i < n) add(idx(i,j,1),idx(i+1,j,0),k,0);
91         if(j < n) add(idx(i,j,1),idx(i,j+1,0),k,0);
92     }
93 }
94 
95 int main() {
96     build(),MCMF(),printf("%d\n",-minn);
97     return 0;
98 }

 

posted @ 2018-11-27 20:33  白怀潇  阅读(145)  评论(0)  编辑  收藏  举报