题意:~
思路:最小权匹配,跑两遍KM,第二次建边要处理每个工人第一次做工的时间差。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
#define pb push_back
#define Key_value ch[ch[root][1]][0]
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define MS(x,y) memset(x,y,sizeof(x))
const double eps = 1e-8;
const double pi = acos (-1.0);
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int M = 1000100;
const int MAXN = 10010;
const int MAXM = 100010;
const int N = 310;
int nx, ny;
int g[N][N];
int linker[N], lx[N], ly[N];
int slack[N];
bool visx[N], visy[N];
bool DFS(int x)
{
visx[x] = true;
for(int y = 1; y <= ny; y++)
{
if(visy[y])continue;
int tmp = lx[x] + ly[y] - g[x][y];
if(tmp == 0)
{
visy[y] = true;
if(linker[y] == -1 || DFS(linker[y]))
{
linker[y] = x;
return true;
}
}
else if(slack[y] > tmp)
slack[y] = tmp;
}
return false;
}
int KM()
{
memset(linker, -1, sizeof(linker));
memset(ly, 0, sizeof(ly));
for(int i = 1; i <= nx; i++)
{
lx[i] = -INF;
for(int j = 1; j <= ny; j++)
if(g[i][j] > lx[i])
lx[i] = g[i][j];
}
for(int x = 1; x <= nx; x++)
{
for(int i = 1; i <= ny; i++)
slack[i] = INF;
while(true)
{
memset(visx, false, sizeof(visx));
memset(visy, false, sizeof(visy));
if(DFS(x))break;
int d = INF;
for(int i = 1; i <= ny; i++)
if(!visy[i] && d > slack[i])
d = slack[i];
for(int i = 1; i <= nx; i++)
if(visx[i])
lx[i] -= d;
for(int i = 1; i <= ny; i++)
{
if(visy[i])ly[i] += d;
else slack[i] -= d;
}
}
}
int res = 0;
for(int i = 1; i <= ny; i++)
{
if(linker[i] != -1)
res += g[linker[i]][i];
}
return res;
}
struct mmp
{
int a, b, c;
} wk[55];
int kk[50];
int mp[50][50];
int main()
{
int n;
int casee = 1;
while(~scanf("%d", &n), n)
{
int tmp;
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= n; ++j)
{
scanf("%d", &tmp);
g[i][j] = -tmp;
}
}
nx = ny = n;
KM();
for(int i = 1; i <= n; ++i)
{
wk[linker[i]].a = i;
wk[linker[i]].c = -g[linker[i]][i];
}
for(int i = 1; i <= n; ++i)
{
kk[i] = -g[linker[i]][i];
}
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= n; ++j)
{
scanf("%d", &tmp);
mp[i][j] = tmp;
if(kk[j] - wk[i].c > 0)
tmp += kk[j] - wk[i].c;
g[i][j] = -tmp;
}
}
KM();
for(int i = 1; i <= n; ++i)
{
wk[linker[i]].b = i;
wk[linker[i]].c += -g[linker[i]][i];
}
int time = 0;
for(int i = 1; i <= n; ++i)
{
time += abs(mp[i][wk[i].b] + g[i][wk[i].b]);
}
printf("Case %d:\n", casee++);
for(int i = 1; i <= n; ++i)
{
printf("Worker %d: %d %d %d\n", i, wk[i].a, wk[i].b, wk[i].c);
}
printf("Total idle time: %d\n", time);
}
return 0;
}
/*
4
8 6 12 19
13 2 18 10
9 15 16 17
5 18 4 10
2 6 3 3
8 5 9 2
5 8 4 3
4 4 5 2
0
*/
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