题意:
1、一个人从[1,1] ->[n,n] ->[1,1]
2、仅仅能走最短路
3、走过的点不能再走
问最大和。
对每一个点拆点限流为1就可以满足3.
费用流流量为2满足1
最大费用流,先给图取负,结果再取负,满足2
#include <stdio.h> #include <string.h> #include <iostream> #include <math.h> #include <queue> #include <set> #include <algorithm> #include <stdlib.h> #define N 605*605*2 #define M N*2 #define inf 1<<29 #define ll int using namespace std; //双向边,注意RE //注意 点标必须是 [0 - 汇点] struct Edge{ ll from, to, flow, cap, nex, cost; }edge[M*2]; ll head[N], edgenum; void add(ll u,ll v,ll cap,ll cost){//网络流要加反向弧 Edge E={u, v, 0, cap, head[u], cost}; edge[edgenum]=E; head[u]=edgenum++; Edge E2={v, u, 0, 0, head[v], -cost}; //这里的cap若是单向边要为0 edge[edgenum]=E2; head[v]=edgenum++; } ll D[N], P[N], A[N]; bool inq[N]; bool BellmanFord(ll s, ll t, ll &flow, ll &cost){ for(ll i=0;i<=t;i++) D[i]= inf; memset(inq, 0, sizeof(inq)); D[s]=0; inq[s]=1; P[s]=0; A[s]=inf; queue<ll> Q; Q.push( s ); while( !Q.empty()){ ll u = Q.front(); Q.pop(); inq[u]=0; for(ll i=head[u]; i!=-1; i=edge[i].nex){ Edge &E = edge[i]; if(E.cap > E.flow && D[E.to] > D[u] +E.cost){ D[E.to] = D[u] + E.cost ; P[E.to] = i; A[E.to] = min(A[u], E.cap - E.flow); if(!inq[E.to]) Q.push(E.to) , inq[E.to] = 1; } } } if(D[t] == inf) return false; flow += A[t]; cost += D[t] * A[t]; ll u = t; while(u != s){ edge[P[u]].flow += A[t]; edge[P[u]^1].flow -= A[t]; u = edge[P[u]].from; } return true; } ll Mincost(ll s,ll t){//返回最小费用 ll flow = 0, cost = 0; while(BellmanFord(s, t, flow, cost)); return cost; } void init(){memset(head,-1,sizeof head); edgenum = 0;} ll n; ll Hash(ll x,ll y){return (x-1)*n+y;} ll Hash2(ll x,ll y){return n*n+(x-1)*n+y;} ll mp[605][605]; int main(){ ll i, j, u, v, cost; while(~scanf("%d",&n)){ init(); for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&mp[i][j]); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { add(Hash(i,j),Hash2(i,j),1,-mp[i][j]); u = Hash2(i,j); if(i!=n) { v = Hash(i+1,j); add(u,v,3,0); } if(j!=n) { v = Hash(i,j+1); add(u,v,3,0); } } } add(Hash(1,1), Hash2(1,1), 1, 0); add(Hash(n,n), Hash2(n,n), 1, 0); printf("%d\n",-Mincost(Hash(1,1), Hash2(n,n))); } return 0; }
时间: 2024-11-05 12:19:04