HPU 2686--Matrix【最大费用最大流 && 经典建图】

Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2062    Accepted Submission(s): 1074

Problem Description

Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.
Every time yifenfei should to do is that choose a detour which frome the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix yifenfei choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And yifenfei can not pass the same area of the Matrix except the start and end.

 

Input

The input contains multiple test cases.
Each case first line given the integer n (2<n<30)
Than n lines,each line include n positive integers.(<100)

 

Output

For each test case output the maximal values yifenfei can get.

 

Sample Input

   
   
   
   

    
    
    
    2
10 3
5 10
3
10 3 3
2 5 3
6 7 10
5
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    28
46
80
   
   
   
   

 

和HPU3376一模一样， 但是数据比较弱，果断水了。

具体的解析请看这里 ：HPU3376

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define INF 0x3f3f3f3f
#define maxn 800000 + 1000
#define maxm 4000000 + 1000
using namespace std;
int n;
int outset;
int inset;
struct node {
    int u, v, cap, flow, cost, next;
};

node edge[maxm];
int head[maxn], cnt;
int per[maxn];
int dist[maxn], vis[maxn];
int map[660][660];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w, int c){
    node E1 = {u, v, w, 0, c, head[u]};
    edge[cnt] = E1;
    head[u] = cnt++;
    node E2 = {v, u, 0, 0, -c, head[v]};
    edge[cnt] = E2;
    head[v] = cnt++;
}

int change(int x, int y){
    return (x - 1) * n + y;
}

void getmap(){
    int t = n * n;
    outset = 0;
    inset = n * n * 2 + 1;
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= n; ++j){
            scanf("%d", &map[i][j]);
            if(i == 1 && j == 1 || i == n && j == n)
                add(change(i, j), change(i, j) + t, 2, map[i][j]);
            else
                add(change(i, j), change(i, j) + t, 1, map[i][j]);
            if(i + 1 <= n)
                add(change(i, j) + t, change(i + 1, j), 1, 0);
            if(j + 1 <= n)
                add(change(i, j) + t, change(i, j + 1), 1, 0);
        }
    }
    add(outset, 1, 2, 0);
    add(change(n, n) + t, inset, 2, 0);
}

bool SPFA(int st, int ed){
    queue<int>q;
    for(int i = 0; i <= inset; ++i){
        dist[i] = -INF;
        vis[i] = 0;
        per[i] = -1;
    }
    dist[st] = 0;
    vis[st] = 1;
    q.push(st);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].next){
            node E = edge[i];
            if(dist[E.v] < dist[u] + E.cost && E.cap > E.flow){
                dist[E.v] = dist[u] + E.cost;
                per[E.v] = i;
                if(!vis[E.v]){
                    vis[E.v] = 1;
                    q.push(E.v);
                }
            }
        }
    }
    return per[ed] != -1;
}

void MCMF(int st, int ed, int &cost, int &flow){
    flow = 0;
    cost = 0;
    while(SPFA(st, ed)){
        int mins = INF;
        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){
            mins = min(mins, edge[i].cap - edge[i].flow);
        }
        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){
            edge[i].flow += mins;
            edge[i ^ 1].flow -= mins;
            cost += edge[i].cost * mins;
        }
        flow += mins;
    }
}
int main (){
    while(scanf("%d", &n) != EOF){
        init();
        getmap();
        int cost, flow;
        MCMF(outset, inset, cost, flow);
        cost = cost - map[1][1] - map[n][n];
        printf("%d\n", cost);
    }
    return 0;
}