Golden Eggs HDU - 3820(最小割)
Golden Eggs
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 673 Accepted Submission(s): 400
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
2 2 100 100
1 1
5 1
1 4
1 1
1 4 85 95
100 100 10 10
10 10 100 100
Case 2: 225
#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 + , INF = 0x7fffffff;
int dir[][] = {{, },{-, },{, },{, -}}; int n, m, G, S, s, t; int head[maxn], cur[maxn], d[maxn], vis[maxn], nex[maxn << ], cnt; struct node
{
int u, v, c;
}Node[maxn << ]; 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, );
} bool bfs()
{
mem(d, );
queue<int> Q;
d[s] = ;
Q.push(s);
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i = head[u]; i != -; i = nex[i])
{
int v = Node[i].v;
if(!d[v] && Node[i].c > )
{
d[v] = d[u] + ;
Q.push(v);
if(v == t) return ;
}
}
}
return d[t] != ;
} int dfs(int u, int cap)
{
int ret = ;
if(u == t || cap == )
return cap;
for(int &i = cur[u]; i != -; i = nex[i])
{
int v = Node[i].v;
if(d[v] == d[u] + && Node[i].c > )
{
int V = dfs(v, min(cap, Node[i].c));
Node[i].c -= V;
Node[i ^ ].c += V;
ret += V;
cap -= V;
if(cap == ) break;
}
}
if(cap > ) d[u] = -;
return ret;
} int Dinic()
{
int ans = ;
while(bfs())
{
memcpy(cur, head, sizeof(head));
ans += dfs(s, INF);
}
return ans;
} int main()
{
int T, kase = ;
rd(T);
while(T--)
{
mem(head, -);
cnt = ;
int w;
rd(n), rd(m), rd(G), rd(S);
s = , t = n * m * + ;
int sum = ;
rap(i, , n)
{
rap(j, , m)
{
rep(k, , )
{
int nx = i + dir[k][];
int ny = j + dir[k][];
if(nx < || ny < || nx > n || ny > m) continue;
add((i - ) * m + j, n * m + (nx - ) * m + ny, (((i + j) & ) ? G : S));
}
rd(w);
sum += w;
if((i + j) & ) add(s, (i - ) * m + j, w);
else add(n * m + (i - ) * m + j, t, w);
add((i - ) * m + j, n * m + (i - ) * m + j, INF);
}
}
rap(i, , n)
rap(j, , m)
{
rd(w);
sum += w;
if((i + j) & ) add(n * m + (i - ) * m + j, t, w);
else add(s, (i - ) * m + j, w);
}
printf("Case %d: ", ++kase);
cout << sum - Dinic() << endl; } return ;
}
Golden Eggs
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 673 Accepted Submission(s): 400
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
2 2 100 100
1 1
5 1
1 4
1 1
1 4 85 95
100 100 10 10
10 10 100 100
Case 2: 225
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