最小K路径覆盖问题。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 205
#define maxm 2000005
#define eps 1e-10
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head
struct Edge
{
int v, c, w, next;
Edge() {}
Edge(int v, int c, int w, int next) : v(v), c(c), w(w), next(next) {}
}E[maxm];
queue<int> q;
char g[60][60];
int H[maxn], cntE;
int dis[maxn];
int cur[maxn];
int vis[maxn];
int cap[maxn];
int flow, cost, T;
int n, m, k, o, s, t, x;
void addedges(int u, int v, int c, int w)
{
E[cntE] = Edge(v, c, w, H[u]);
H[u] = cntE++;
E[cntE] = Edge(u, 0, -w, H[v]);
H[v] = cntE++;
}
bool spfa(void)
{
memset(dis, INF, sizeof dis);
vis[s] = ++T;
cap[s] = INF;
dis[s] = 0;
cur[s] = -1;
q.push(s);
while(!q.empty()) {
int u = q.front();
q.pop();
vis[u] = T - 1;
for(int e = H[u]; ~e; e = E[e].next) {
int v = E[e].v, c = E[e].c, w = E[e].w;
if(c && dis[v] > dis[u] + w) {
dis[v] = dis[u] + w;
cap[v] = min(c, cap[u]);
cur[v] = e;
if(vis[v] != T) {
vis[v] = T;
q.push(v);
}
}
}
}
if(dis[t] == INF) return false;
flow += cap[t];
cost += cap[t] * dis[t];
for(int e = cur[t]; ~e; e = cur[E[e^1].v]) {
E[e].c -= cap[t];
E[e ^ 1].c += cap[t];
}
}
void mcmf(void)
{
flow = cost = 0;
while(spfa());
}
void init(void)
{
cntE = T = 0;
memset(H, -1, sizeof H);
}
void read(void)
{
scanf("%d%d%d", &n, &m, &k);
for(int i = 1; i <= n; i++) scanf("%s", g[i] + 1);
}
inline int calc(int i, int j)
{
return (i - 1) * m + j;
}
void work(void)
{
o = n * m, s = 2 * o + 1, t = 2 * o + 2, x = 2 * o + 3;
addedges(s, x, k, 0);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++) {
addedges(s, calc(i, j), 1, 0);
addedges(calc(i, j) + o, t, 1, 0);
addedges(x, calc(i, j) + o, 1, 0);
for(int k = i + 1; k <= n; k++) {
int t = k - i - 1;
if(g[i][j] == g[k][j]) t -= g[i][j] - '0';
addedges(calc(i, j), calc(k, j) + o, 1, t);
}
for(int k = j + 1; k <= m; k++) {
int t = k - j - 1;
if(g[i][j] == g[i][k]) t -= g[i][j] - '0';
addedges(calc(i, j), calc(i, k) + o, 1, t);
}
}
mcmf();
if(flow != o) printf("-1\n");
else printf("%d\n", -cost);
}
int main(void)
{
int _, __;
while(scanf("%d", &_)!=EOF) {
__ = 0;
while(_--) {
init();
read();
printf("Case %d : ", ++__);
work();
}
}
return 0;
}
