方格取数(2)
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6806 Accepted Submission(s): 2175
Problem Description
给你一个m*n的格子的棋盘，每个格子里面有一个非负数。
从中取出若干个数，使得任意的两个数所在的格子没有公共边，就是说所取数所在的2个格子不能相邻，并且取出的数的和最大。
从中取出若干个数，使得任意的两个数所在的格子没有公共边，就是说所取数所在的2个格子不能相邻，并且取出的数的和最大。
Input
包括多个测试实例，每个测试实例包括2整数m,n和m*n个非负数(m<=50,n<=50)
Output
对于每个测试实例，输出可能取得的最大的和
Sample Input
3 3
75 15 21
75 15 28
34 70 5
Sample Output
188
Author
ailyanlu
Source
Recommend
8600
析：很明显的是二分图的最大独立集，但是每个点都有权值，这个可以用最小割来求，建立一个超级源点s，和汇点t，然后s 向 X集，添加容量为权值的边，Y集向 t 添加容量为权值的，然后跑一遍最小割，然后用总权值减去就是答案了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50 * 50 + 20;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
void init(int n){
this-> n = n;
for(int i = 0; i < n; ++i) G[i].cl;
edges.cl;
}
void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m-2);
G[to].pb(m-1);
}
bool bfs(){
queue<int> q;
ms(vis, 0); d[s] = 0;
q.push(s); vis[s] = 1;
while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i < G[x].sz; ++i){
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow){
vis[e.to] = 1;
d[e.to] = d[x] + 1;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f = 0;
for(int &i = cur[x]; i < G[x].sz; ++i){
Edge &e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int maxflow(int s, int t){
this->s = s;
this->t = t;
int flow = 0;
while(bfs()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
}
};
Dinic dinic;
int main(){
while(scanf("%d %d", &n, &m) == 2){
dinic.init(n * m + 10);
int s = 0, t = n * m + 5;
int sum = 0;
FOR(i, 0, n) for(int j = 1; j <= m; ++j){
int x;
scanf("%d", &x);
sum += x;
int now = i * m + j;
if(i + j & 1){
dinic.addEdge(s, now, x);
if(i) dinic.addEdge(now, now - m, INF); // up
if(j > 1) dinic.addEdge(now, now - 1, INF); // left
if(i + 1 < n) dinic.addEdge(now, now + m, INF); // down
if(j < m) dinic.addEdge(now, now + 1, INF); // right
}
else dinic.addEdge(now, t, x);
}
printf("%d\n", sum - dinic.maxflow(s, t));
}
return 0;
}
