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 400010const int inf = 0x3f3f3f3f;struct E{    int to, frm, nxt, cap;}edge[M];int head[N],e,n,m,src,des;int dep[N], gap[N];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[N];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[N],stack[N];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;}