题目:给出n*n的方格矩阵,现在从左上方走m次到右下方,问m次能够获得的最大价值和。
分析:最大费用流。拆点进行限制每个格子只取一次,假设点x拆成 x,xx,右边(假设有)y,yy,下方(假设有)z,zz
点 点 流量 费用
则:x , xx , 1 , -a[i][j]
x , xx , m , 0
xx, y , m , 0
xx , z , m , 0
用最小费用流增广m次即可
#include <set> #include <map> #include <list> #include <cmath> #include <queue> #include <stack> #include <string> #include <vector> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; #define debug puts("here") #define rep(i,n) for(int i=0;i<n;i++) #define rep1(i,n) for(int i=1;i<=n;i++) #define REP(i,a,b) for(int i=a;i<=b;i++) #define foreach(i,vec) for(unsigned i=0;i<vec.size();i++) #define pb push_back #define RD(n) scanf("%d",&n) #define RD2(x,y) scanf("%d%d",&x,&y) #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w) #define All(vec) vec.begin(),vec.end() #define MP make_pair #define PII pair<int,int> #define PQ priority_queue #define cmax(x,y) x = max(x,y) #define cmin(x,y) x = min(x,y) #define Clear(x) memset(x,0,sizeof(x)) /* #pragma comment(linker, "/STACK:1024000000,1024000000") int size = 256 << 20; // 256MB char *p = (char*)malloc(size) + size; __asm__("movl %0, %%esp " :: "r"(p) ); */ /******** program ********************/ char op; inline void Int(int &x){ while( !isdigit(op=getchar()) ); x = op-'0'; while(isdigit(op=getchar())) x = x*10+op-'0'; } const int MAXN = 5005; const int MAXM = 200005; const int INF = 1e9; int pre[MAXN],dis[MAXN]; int po[MAXN],tol; bool use[MAXN]; int q[MAXM],head,tail; int n,m,vs,vt,ans; int a[55][55]; struct node{ int y,f,cost,next; }edge[MAXM]; inline void Add(int x,int y,int f,int cost){ edge[++tol].y = y; edge[tol].f = f; edge[tol].cost = cost; edge[tol].next = po[x]; po[x] = tol; } inline void add(int x,int y,int f,int cost){ Add(x,y,f,cost); Add(y,x,0,-cost); } inline bool spfa(){ memset(use,false,sizeof(use)); rep1(i,vt) dis[i] = INF; dis[vs] = 0; head = tail = 0; q[tail++] = vs; pre[vs] = 0; while(head<tail){ int x = q[head++]; use[x] = false; for(int i=po[x];i;i=edge[i].next){ int y = edge[i].y; if(edge[i].f>0&&edge[i].cost+dis[x]<dis[y]){ dis[y] = dis[x]+edge[i].cost; pre[y] = i; if(!use[y]){ use[y] = true; q[tail++] = y; } } } } if(dis[vt]==INF) return false; int aug = INF; for(int i=pre[vt];i;i=pre[edge[i^1].y]) aug = min(aug,edge[i].f); for(int i=pre[vt];i;i=pre[edge[i^1].y]){ edge[i].f -= aug; edge[i^1].f += aug; } ans += dis[vt]*aug; return true; } inline int id(int x,int y){ return (x-1)*n+y; } int main(){ #ifndef ONLINE_JUDGE freopen("sum.in","r",stdin); //freopen("sum.out","w",stdout); #endif while(~RD2(n,m)){ int nn = n*n; vs = 1; vt = nn<<1; tol = 1; Clear(po); rep1(i,n) rep1(j,n) Int(a[i][j]); //RD(a[i][j]); rep1(i,n){ rep1(j,n){ int now = id(i,j); add( now,now+nn,1,-a[i][j] ); add( now,now+nn,m,0 ); if(i<n) add( now+nn,now+n,m,0 ); if(j<n) add( now+nn,now+1,m,0 ); } } ans = 0; while(m--&&spfa()) ; cout<<-ans<<endl; } return 0; }