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\n", 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 = ;
const int INF = <<;

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, , -cost, G[from].size()-});
}

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

        while (!que.empty()) {
            int u = que.front(); que.pop(); in[u] = ;
            for (int i = ; i < G[u].size(); ++i) {
                Edge &e = G[u][i];
                if (e.cap >  && 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] == ) {
                        in[e.to] = ;
                        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+; }
int getid2(int x, int y) { return x*n+y+n*n+; }
int a[][];
int main()
{
    while (~scanf("%d", &n)) {
        for (int i = ; i < n; ++i) {
            for (int j = ; j < n; ++j) {
                scanf("%d", &a[i][j]);
            }
        }
        for (int i = ; i <= *n*n; ++i) G[i].clear();
        for (int i = ; i < n; ++i) {
            for (int j = ; j < n; ++j) {
                if (i+j == ) addedge(getid1(i,j), getid2(i,j), , -a[i][j]);
                else if (i+j == n* - ) addedge(getid1(i,j), getid2(i, j), , -a[i][j]);
                else addedge(getid1(i, j), getid2(i, j), , -a[i][j]);
                if (i+<n) addedge(getid2(i, j), getid1(i+, j), , );
                if (j+<n) addedge(getid2(i, j), getid1(i, j+), , );
            }
        }
        printf("%d\n", -min_cost_max_flow(getid1(,), getid2(n-,n-)) - a[][] - a[n-][n-]);
    }
    return ;
}

hdu2686

#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\n", 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 = **+;
const int INF = <<;

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, , -cost, G[from].size()-});
}

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

        while (!que.empty()) {
            int u = que.front(); que.pop(); in[u] = ;
            for (int i = ; i < G[u].size(); ++i) {
                Edge &e = G[u][i];
                if (e.cap >  && 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] == ) {
                        in[e.to] = ;
                        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 = ;
    while (ch < '' || ch > '') ch = getchar();
    do {
        data = data* + ch-'';
        ch = getchar();
    } while (ch >= '' && ch <= '');
    return data;
}

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

hdu3376