HPU 2686 -- Matrix [maximum cost flow & amp; Classic Graph creation], hpu2686 -- matrix

HPU 2686 -- Matrix [maximum cost flow & Classic Graph creation], hpu2686 -- matrix

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


Problem DescriptionYifenfei 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 shocould 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 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 before t the start and end.
 
InputThe 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)
 
OutputFor each test case output the maximal values yifenfei can get.
Sample Input

210 35 10310 3 32 5 36 7 1051 2 3 4 52 3 4 5 63 4 5 6 74 5 6 7 85 6 7 8 9

 
Sample Output

284680

 

It is exactly the same as HPU3376, but the data is weak and decisive.

For details, see HPU3376.

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define INF 0x3f3f3f3f#define maxn 800000 + 1000#define maxm 4000000 + 1000using 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;}

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