HDU 3376 && 2686 方格取数 最大和 费用流裸题

题意：

1、一个人从[1,1] ->[n,n] ->[1,1] 

2、只能走最短路

3、走过的点不能再走

问最大和。

对每个点拆点限流为1即可满足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 intusing 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;}