HDU2686-Matrix & HDU3376-Matrix Again（费用流）

比较简单的题了。

只需从左上角到右下角找两条路就可以了。

因为每个点只能走一次，所以拆点，限制流量为1。

因为求的是最大值，所以权值取反求最小值。

因为第一个点和最后一个点经过两次，只算一次，最后要减去。

ps：数组还是开大点好。。。。不知道什么时候就SB了。。。

注意汇点可能不是最后一个点（模板的问题。。

注意初始化的范围。。。。

matrix again只是数据大了，算法不用改。。。无聊。。。。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <bitset>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <map>
#include <set>
#define pk(x) printf("%d
", x)
using namespace std;
#define PI acos(-1.0)
#define EPS 1E-6
#define clr(x,c) memset(x,c,sizeof(x))
//#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;

const int MAXV = 2000;
const int INF = 1<<30;

struct Edge { int to, cap, cost, rev; };
vector<Edge> G[MAXV];
int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV];
queue<int> que;

void addedge(int from, int to, int cap, int cost) {
    G[from].push_back((Edge){to, cap, cost, G[to].size()});
    G[to].push_back((Edge){from, 0, -cost, G[from].size()-1});
}

int min_cost_max_flow(int s, int t) { //, int f) {
    int res = 0;
    int f = 0;
    while (1) { //f > 0) {
        for (int i = 1; i <= t; ++i) dist[i] = INF, in[i] = 0;
        dist[s] = 0;
        while (!que.empty()) que.pop();
        in[s] = 1;
        que.push(s);

        while (!que.empty()) {
            int u = que.front(); que.pop(); in[u] = 0;
            for (int i = 0; i < G[u].size(); ++i) {
                Edge &e = G[u][i];
                if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {
                    dist[e.to] = dist[u] + e.cost;

                    prv[e.to] = u;
                    pre[e.to] = i;
                    if (in[e.to] == 0) {
                        in[e.to] = 1;
                        que.push(e.to);
                    }
                }
            }
        }

        if (dist[t] == INF) break; //return -1;

        int d = INF; // d = f;
        for (int v = t; v != s; v = prv[v]) {
            d = min(d, G[prv[v]][pre[v]].cap);
        }
        f += d;
        res += d * dist[t];
        for (int v = t; v != s; v = prv[v]) {
            Edge &e = G[prv[v]][pre[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
    //return f;
}

int n;
int getid1(int x, int y) { return x*n+y+1; }
int getid2(int x, int y) { return x*n+y+n*n+1; }
int a[30][30];
int main()
{
    while (~scanf("%d", &n)) {
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                scanf("%d", &a[i][j]);
            }
        }
        for (int i = 0; i <= 2*n*n; ++i) G[i].clear();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i+j == 0) addedge(getid1(i,j), getid2(i,j), 2, -a[i][j]);
                else if (i+j == n*2 - 2) addedge(getid1(i,j), getid2(i, j), 2, -a[i][j]);
                else addedge(getid1(i, j), getid2(i, j), 1, -a[i][j]);
                if (i+1<n) addedge(getid2(i, j), getid1(i+1, j), 1, 0);
                if (j+1<n) addedge(getid2(i, j), getid1(i, j+1), 1, 0);
            }
        }
        printf("%d
", -min_cost_max_flow(getid1(0,0), getid2(n-1,n-1)) - a[0][0] - a[n-1][n-1]);
    }
    return 0;
}

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <bitset>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <map>
#include <set>
#define pk(x) printf("%d
", x)
using namespace std;
#define PI acos(-1.0)
#define EPS 1E-6
#define clr(x,c) memset(x,c,sizeof(x))
//#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;

const int MAXV = 610*610*2+2;
const int INF = 1<<30;

struct Edge { int to, cap, cost, rev; };
vector<Edge> G[MAXV];
int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV];
queue<int> que;

void addedge(int from, int to, int cap, int cost) {
    G[from].push_back((Edge){to, cap, cost, G[to].size()});
    G[to].push_back((Edge){from, 0, -cost, G[from].size()-1});
}

int min_cost_max_flow(int s, int t) { //, int f) {
    int res = 0;
    int f = 0;
    while (1) { //f > 0) {
        for (int i = 1; i <= t; ++i) dist[i] = INF, in[i] = 0;
        dist[s] = 0;
        while (!que.empty()) que.pop();
        in[s] = 1;
        que.push(s);

        while (!que.empty()) {
            int u = que.front(); que.pop(); in[u] = 0;
            for (int i = 0; i < G[u].size(); ++i) {
                Edge &e = G[u][i];
                if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {
                    dist[e.to] = dist[u] + e.cost;

                    prv[e.to] = u;
                    pre[e.to] = i;
                    if (in[e.to] == 0) {
                        in[e.to] = 1;
                        que.push(e.to);
                    }
                }
            }
        }

        if (dist[t] == INF) break; //return -1;

        int d = INF; // d = f;
        for (int v = t; v != s; v = prv[v]) {
            d = min(d, G[prv[v]][pre[v]].cap);
        }
        f += d;
        res += d * dist[t];
        for (int v = t; v != s; v = prv[v]) {
            Edge &e = G[prv[v]][pre[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
    //return f;
}

inline int Scan()
{
    char ch = getchar();
    int data = 0;
    while (ch < '0' || ch > '9') ch = getchar();
    do {
        data = data*10 + ch-'0';
        ch = getchar();
    } while (ch >= '0' && ch <= '9');
    return data;
}

int n;
int getid1(int x, int y) { return x*n+y+1; }
int getid2(int x, int y) { return x*n+y+n*n+1; }
int a[600][600];
int main()
{
    while (~scanf("%d", &n)) {
        int maxn = 2*n*n;
        for (int i = 0; i <= maxn; ++i) G[i].clear();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                a[i][j] = Scan();
                if (i+j == 0 || i+j == n*2-2) addedge(getid1(i,j), getid2(i,j), 2, -a[i][j]);
                else addedge(getid1(i, j), getid2(i, j), 1, -a[i][j]);
                if (i+1<n) addedge(getid2(i, j), getid1(i+1, j), 1, 0);
                if (j+1<n) addedge(getid2(i, j), getid1(i, j+1), 1, 0);
            }
        }
        printf("%d
", -min_cost_max_flow(getid1(0,0), getid2(n-1,n-1)) - a[0][0] - a[n-1][n-1]);
    }
    return 0;
}