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;
}

 

来源：https://www.cnblogs.com/dashuzhilin/p/4655792.html