HDU 4862 Jump（最大k路径覆盖 费用流）

题意：一个n*m的矩阵，需要遍历所有点，从起点出发每次只可向右或向下跳，若到达位置的数字与上一步的数字相同，则获得该数字大小的能量；

否则消耗能量：哈密顿距离减1；求可获得的最大能量；

思路：网络流之最大k路径覆盖。

源点向n*m（X图）各点建流量为1，费用为0的边；

n*m（Y图）各点向汇点建流量为1，费用为0的边；

新增一个起点；

源点向起点建流量为k,费用为0的边；起点向各点建流量1，费用为0的边；

n*m各点间建边；

建好图后跑最小费用最大流，如果满流则存在解，否则不存在；最小费用的相反数就是所能够获得的最大能量；

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 1010;
const int MAXM = 10010;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;
void init(int n)
{
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i = 0;i < N;i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u];i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(edge[i].cap > edge[i].flow &&
dis[v] > dis[u] +edge[i].cost)
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
int minCostMaxflow(int s,int t,int &cost)
{
int flow = 0;
cost = 0;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t];i != -1 ;i = pre[edge[i^1].to])
{
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost*Min;
}
flow += Min;
}
return flow;
}
int n,m,k;
char str[12][12];
void solve()
{
init(2*n*m + 3);
int start = 2*n*m;
int end = 2*n*m+2;
addedge(start,start+1,k,0);
for(int i = 0;i < n;i++)
for(int j = 0;j < m;j++)
{
addedge(start,2*(i*m+j),1,0);
addedge(2*(i*m+j)+1,end,1,0);
addedge(start+1,2*(i*m+j)+1,1,0);
for(int y = j+1;y < m;y++) //向右跳
{
if(str[i][y] == str[i][j])
addedge(2*(i*m+j),2*(i*m+y)+1,1,-(str[i][j]-'0')+y-j-1);
else addedge(2*(i*m+j),2*(i*m+y)+1,1,y-j-1);
}
for(int x = i+1; x < n;x++)//向下跳
{
if(str[x][j] == str[i][j])
addedge(2*(i*m+j),2*(x*m+j)+1,1,-(str[i][j]-'0')+x-i-1);
else addedge(2*(i*m+j),2*(x*m+j)+1,1,x-i-1);
}
}
int cost;
int flow = minCostMaxflow(start,end,cost);
if(flow != n*m)printf("-1\n");
else printf("%d\n",-cost);
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
int iCase = 0;
while(T--)
{
iCase++;
scanf("%d%d%d",&n,&m,&k);
for(int i = 0;i < n;i++)
scanf("%s",str[i]);
printf("Case %d : ",iCase);
solve();
}
return 0;
}