Golden Eggs

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 673    Accepted Submission(s): 400

Problem Description There is a grid with N rows and M columns. In each cell you can choose to put a golden or silver egg in it, or just leave it empty. If you put an egg in the cell, you will get some points which depends on the color of the egg. But for every pair of adjacent eggs with the same color, you lose G points if there are golden and lose S points otherwise. Two eggs are adjacent if and only if there are in the two cells which share an edge. Try to make your points as high as possible.  

 

Input The first line contains an integer T indicating the number of test cases.
There are four integers N, M, G and S in the first line of each test case. Then 2*N lines follows, each line contains M integers. The j-th integer of the i-th line Aij indicates the points you will get if there is a golden egg in the cell(i,j). The j-th integer of the (i+N)-th line Bij indicates the points you will get if there is a silver egg in the cell(i,j).

Technical Specification
1. 1 <= T <= 20
2. 1 <= N,M <= 50
3. 1 <= G,S <= 10000
4. 1 <= Aij,Bij <= 10000  

 

Output For each test case, output the case number first and then output the highest points in a line.  

 

Sample Input 2 2 2 100 100 1 1 5 1 1 4 1 1 1 4 85 95 100 100 10 10 10 10 100 100  

 

Sample Output Case 1: 9 Case 2: 225  

 

Author hanshuai  

 

Source The 6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Final   看到這種棋盤問題求最大值 那就是黑白染色最小割 把每個點拆成u  v  白點 s 向 u 連一條權值為 w(當前點放金蛋) 的邊   u 向 v連一條權值為INF 的邊  v向t連一條權值為 w’(當前點放銀蛋) 的邊  黑點 與之相反 然後黑點的u向相鄰白點的v連一條權值為G的邊 白點的u向黑點的u連一條權值為S的邊 跑最小割即可

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + 10, INF = 0x7fffffff;
int dir[4][2] = {{1, 0},{-1, 0},{0, 1},{0, -1}};

int n, m, G, S, s, t;

int head[maxn], cur[maxn], d[maxn], vis[maxn], nex[maxn << 1], cnt;

struct node
{
    int u, v, c;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    mem(d, 0);
    queue<int> Q;
    d[s] = 1;
    Q.push(s);
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = nex[i])
        {
            int v = Node[i].v;
            if(!d[v] && Node[i].c > 0)
            {
                d[v] = d[u] + 1;
                Q.push(v);
                if(v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(d[v] == d[u] + 1 && Node[i].c > 0)
        {
            int V = dfs(v, min(cap, Node[i].c));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

int Dinic()
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        ans += dfs(s, INF);
    }
    return ans;
}

int main()
{
    int T, kase = 0;
    rd(T);
    while(T--)
    {
        mem(head, -1);
        cnt = 0;
        int w;
        rd(n), rd(m), rd(G), rd(S);
        s = 0, t = n * m * 2 + 10;
        int sum = 0;
        rap(i, 1, n)
        {
            rap(j, 1, m)
            {
                rep(k, 0, 4)
                {
                    int nx = i + dir[k][0];
                    int ny = j + dir[k][1];
                    if(nx < 1 || ny < 1 || nx > n || ny > m) continue;
                    add((i - 1) * m + j, n * m + (nx - 1) * m + ny, (((i + j) & 1) ? G : S));
                }
                rd(w);
                sum += w;
                if((i + j) & 1) add(s, (i - 1) * m + j, w);
                else add(n * m + (i - 1) * m + j, t, w);
                add((i - 1) * m + j, n * m + (i - 1) * m + j, INF);
            }
        }
        rap(i, 1, n)
            rap(j, 1, m)
            {
                rd(w);
                sum += w;
                if((i + j) & 1) add(n * m + (i - 1) * m + j, t, w);
                else add(s, (i - 1) * m + j, w);
            }
        printf("Case %d: ", ++kase);
        cout << sum - Dinic() << endl;


    }


    return 0;
}

 

Golden Eggs

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 673    Accepted Submission(s): 400

Problem Description There is a grid with N rows and M columns. In each cell you can choose to put a golden or silver egg in it, or just leave it empty. If you put an egg in the cell, you will get some points which depends on the color of the egg. But for every pair of adjacent eggs with the same color, you lose G points if there are golden and lose S points otherwise. Two eggs are adjacent if and only if there are in the two cells which share an edge. Try to make your points as high as possible.  

 

Input The first line contains an integer T indicating the number of test cases.
There are four integers N, M, G and S in the first line of each test case. Then 2*N lines follows, each line contains M integers. The j-th integer of the i-th line Aij indicates the points you will get if there is a golden egg in the cell(i,j). The j-th integer of the (i+N)-th line Bij indicates the points you will get if there is a silver egg in the cell(i,j).

Technical Specification
1. 1 <= T <= 20
2. 1 <= N,M <= 50
3. 1 <= G,S <= 10000
4. 1 <= Aij,Bij <= 10000  

 

Output For each test case, output the case number first and then output the highest points in a line.  

 

Sample Input 2 2 2 100 100 1 1 5 1 1 4 1 1 1 4 85 95 100 100 10 10 10 10 100 100  

 

Sample Output Case 1: 9 Case 2: 225  

 

Author hanshuai  

 

Source The 6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Final