HDU 3376 Matrix Again

Matrix Again

Time Limit: 2000ms

Memory Limit: 102400KB

This problem will be judged on HDU. Original ID: 3376
64-bit integer IO format: %I64d      Java class name: Main

Starvae 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 starvae should to do is that choose a detour which from 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 starvae 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 starvae can not pass the same area of the Matrix except the start and end..
Do you know why call this problem as “Matrix Again”? AS it is like the problem 2686 of HDU.

 

Input

The input contains multiple test cases.
Each case first line given the integer n (2<=n<=600) 
Then n lines, each line include n positive integers. (<100)

 

Output

For each test case output the maximal values starvae 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

Source

HDOJ Monthly Contest – 2010.04.04

 

解题：费用流

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 601*602*2;
struct arc {
    int to,flow,cost,next;
    arc(int x = 0,int y = 0,int z = 0,int nxt = -1) {
        to = x;
        flow = y;
        cost = z;
        next = nxt;
    }
} e[3000010];
int head[maxn],d[maxn],p[maxn],tot,S,T;
bool in[maxn];
void add(int u,int v,int flow,int cost) {
    e[tot] = arc(v,flow,cost,head[u]);
    head[u] = tot++;
    e[tot] = arc(u,0,-cost,head[v]);
    head[v] = tot++;
}
bool spfa() {
    memset(in,false,sizeof in);
    memset(d,-1,sizeof d);
    memset(p,-1,sizeof p);
    queue<int>q;
    d[S] = 0;
    q.push(S);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        in[u] = false;
        for(int i = head[u]; ~i; i = e[i].next) {
            if(e[i].flow && d[e[i].to] < d[u] + e[i].cost) {
                d[e[i].to] = d[u] + e[i].cost;
                p[e[i].to] = i;
                if(!in[e[i].to]) {
                    in[e[i].to] = true;
                    q.push(e[i].to);
                }
            }
        }
    }
    return p[T] > -1;
}
int solve(int ret = 0) {
    while(spfa()) {
        int mh = 1;
        //for(int i = p[T]; ~i; i = p[e[i^1].to])
            //mh = min(mh,e[i].flow);
        for(int i = p[T]; ~i; i = p[e[i^1].to]) {
            e[i].flow -= mh;
            e[i^1].flow += mh;
        }
        ret += mh*d[T];
    }
    return ret;
}
int main() {
    int n;
    while(~scanf("%d",&n)) {
        memset(head,-1,sizeof head);
        int ret = tot = 0;
        for(int i = 1; i <= n; ++i)
            for(int j = 1,val; j <= n; ++j) {
                scanf("%d",&val);
                int u = (i-1)*n + j,v = u + n*n;
                if(u == 1 || u == n*n) {
                    add(u,v,2,val);
                    ret -= val;
                } else add(u,v,1,val);
                if(i < n) add(v,u + n,1,0);
                if(j < n) add(v,u + 1,1,0);
            }
        S = 1;
        T = 2*n*n;
        printf("%d
",solve() + ret);
    }
    return 0;
}