[acm/icpc2016ChinaFinal][CodeforcesGym101194] Mr.Panda and TubeMaster

这个题从范围来看，不难想到跟费用流有关系。但感觉跟费用流联系起来还是很难呀。拆点，这个题可以看成是给每个点找一个后继点，这一点是解题的关键。然后就是要黑白染色定向，定向这个点能走横边还是竖边，定向之后，神奇的发现所有的边都被连了有且仅有一次，这种套路估计大佬已经习以为常了吧。然后对于那些非限制格子，自己的入点连出点，保证可以不连接。然后直接看代码吧。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2000+5;
const int inf = 1e9;
int n,m;
int s,e;

int cnt,head[MAXN];
struct node
{
int u,v,w,f,next;
} edge[50000];

void init()
{
cnt = 0;
for(int i = 0; i <= e; ++i)head[i] = -1;
}

void add(int u,int v,int w,int f)
{
edge[cnt].u = u;
edge[cnt].v = v;
edge[cnt].w = w;
edge[cnt].f = f;
edge[cnt].next = head[u];
head[u] = cnt++;
edge[cnt].u = v;
edge[cnt].v = u;
edge[cnt].w = -w;
edge[cnt].f = 0;
edge[cnt].next = head[v];
head[v] = cnt++;
}

bool vis[MAXN];
int dis[MAXN],pre[MAXN];
bool spfa()
{
for(int i = 0; i <= e; ++i)
{
dis[i] = -inf;
pre[i] = -1;
}
queue<int>q;
q.push(s);
dis[s] = 0;
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = 0;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
int w = edge[i].w;
int f = edge[i].f;
if(f > 0 && dis[v] < dis[u] + w)
{
dis[v] = dis[u] + w;
pre[v] = i;
if(!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
}
if(pre[e] == -1)return 0;
return 1;
}
void get_mincost()
{
int max_flow = 0,min_cost = 0;
while(spfa())
{
int p = pre[e];
int flow = inf;
while(p != -1)
{
flow = min(flow,edge[p].f);
p = pre[edge[p].u];
}
max_flow += flow;
min_cost += flow*dis[e];
p = pre[e];
while(p != -1)
{
edge[p].f -= flow;
edge[p^1].f += flow;
p = pre[edge[p].u];
}
}
if(max_flow != n*m)puts("Impossible");
else printf("%d\n",min_cost);
}

int scorec[40][40],scorer[40][40];
int L[40][40],R[40][40];
bool tu[40][40];

int main()
{
int t,ca = 0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int id = 0;
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)L[i][j] = ++id,R[i][j] = ++id;
s = 0,e = ++id;
for(int i = 0; i < n; ++i)
for(int j = 0; j < m-1; ++j)scanf("%d",&scorec[i][j]);
for(int i = 0; i < n-1; ++i)
for(int j = 0; j < m; ++j)scanf("%d",&scorer[i][j]);
int E;
int x,y;
scanf("%d",&E);
while(E--)
{
scanf("%d%d",&x,&y);
x--,y--;
tu[x][y] = 1;
}
init();
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
{
if((i+j)%2)
{
if(j+1 < m)add(L[i][j],R[i][j+1],scorec[i][j],1);
if(j-1 >= 0)add(L[i][j],R[i][j-1],scorec[i][j-1],1);
}
else
{
if(i+1 < n)add(L[i][j],R[i+1][j],scorer[i][j],1);
if(i-1 >= 0)add(L[i][j],R[i-1][j],scorer[i-1][j],1);
}
}
for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
{
add(s,L[i][j],0,1);
add(R[i][j],e,0,1);
if(!tu[i][j])add(L[i][j],R[i][j],0,1);
else tu[i][j] = 0;
}
printf("Case #%d: ",++ca);
get_mincost();
}
return 0;
}