hdu 1569 方格取数(2) (最大流最小割)

题解：开始想到DP，但数据比较大，用DP会超时。于是将它转化为最小割问题，增加一个源点和一个汇点，把i+j为偶数的节点与源点s相连，权值为棋盘上对应位置的值，其它的节点与汇点t相连，权值还是棋盘上对应位置的值，相邻的节点相连的权值记为无穷大，再根据最大流最小割定理求一遍最大流，答案即为棋盘上所有点的权值-最大流。

代码：

#include <stdio.h>
#include <string.h>
#define N 10010
#define M 400010
const int inf = 0x3f3f3f3f;
struct E
{
int to, frm, nxt, cap;
}edge[M];

int head
,e,n,m,src,des;
int dep
, gap
;

void addedge(int u, int v, int w)
{
edge[e].frm = u;
edge[e].to = v;
edge[e].cap = w;
edge[e].nxt = head[u];
head[u] = e++;
edge[e].frm = v;
edge[e].to = u;
edge[e].cap = 0;
edge[e].nxt = head[v];
head[v] = e++;
}

int que
;

void BFS()
{
memset(dep, -1, sizeof(dep));
memset(gap, 0, sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[des] = 0;
que[rear++] = des;
int u, v;
while (front != rear)
{
u = que[front++];
front = front%N;
for (int i=head[u]; i!=-1; i=edge[i].nxt)
{
v = edge[i].to;
if (edge[i].cap != 0 || dep[v] != -1)
continue;
que[rear++] = v;
rear = rear % N;
++gap[dep[v] = dep[u] + 1];
}
}
}
int cur
,stack
;
int Sap()
{
int res = 0;
BFS();
int top = 0;
memcpy(cur, head, sizeof(head));
int u = src, i;
while (dep[src] < n*m+1)
{
if (u == des)
{
int temp = inf, inser = n;
for (i=0; i!=top; ++i)
if (temp > edge[stack[i]].cap)
{
temp = edge[stack[i]].cap;
inser = i;
}
for (i=0; i!=top; ++i)
{
edge[stack[i]].cap -= temp;
edge[stack[i]^1].cap += temp;
}
res += temp;
top = inser;
u = edge[stack[top]].frm;
}

if (u != des && gap[dep[u] -1] == 0)
break;
for (i = cur[u]; i != -1; i = edge[i].nxt)
if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1)
break;

if (i != -1)
{
cur[u] = i;
stack[top++] = i;
u = edge[i].to;
}
else
{
int min = n*m+3;
for (i = head[u]; i != -1; i = edge[i].nxt)
{
if (edge[i].cap == 0)
continue;
if (min > dep[edge[i].to])
{
min = dep[edge[i].to];
cur[u] = i;
}
}
--gap[dep[u]];
++gap[dep[u] = min + 1];
if (u != src)
u = edge[stack[--top]].frm;
}
}
return res;
}

int main()
{
int i,j,c;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(head,-1,sizeof(head));
src=0,des=n*m+1;
int sum=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
scanf("%d",&c);
sum+=c;
int num=(i-1)*m+j;
if((i+j)%2)
{
addedge(src,num,c);
if(i>1)
addedge(num,num-m,inf);
if(i<n)
addedge(num,num+m,inf);
if(j>1)
addedge(num,num-1,inf);
if(j<m)
addedge(num,num+1,inf);
}
else
addedge(num,des,c);
}
printf("%d\n",sum-Sap());
}
return 0;
}