poj2396 Budget 上下界可行流

Budget：http://poj.org/problem?id=2396

题意：

　　给定一个棋盘，给定每一行每一列的和，还有每个点的性质。求一个合理的棋盘数值放置方式。

思路：

　　比较经典的网络流模型，把每一列看成一个点，每一行看成一个点，利用上下界可行流的思路建图就行了，注意这里由于是严格的小于和大于，所以可以利用 x+1, x-1。

还有就是这道题的0 ， 0 说的是对整张图的操作。

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 1e4 + 9;
int n, m;
struct E {
    int v, val, id;
    int nxt;
} edge[maxn];
int head[maxn], gtot;
void addedge(int u, int v, int val, int id) {
    edge[gtot].v = v;
    edge[gtot].val = val;
    edge[gtot].nxt = head[u];
    edge[gtot].id = -1;
    head[u] = gtot++;

    edge[gtot].v = u;
    edge[gtot].val = 0;
    edge[gtot].nxt = head[v];
    edge[gtot].id = id;
    head[v] = gtot++;
}

int dis[maxn], cur[maxn], all;
bool bfs(int s, int t) {
    memset(dis, inf, sizeof(dis));

    for (int i = 0; i <= all; i++) cur[i] = head[i];
    queue<int> que;
    que.push(s);
    dis[s] = 0;
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v, val = edge[i].val;
            if (val > 0 && dis[v] > dis[u] + 1) {
                dis[v] = dis[u] + 1;
                que.push(v);
            }
        }
    }
    return dis[t] < inf;
}
int dfs(int u, int t, int maxflow) {
    if (u == t || maxflow == 0) return maxflow;
    for (int i = cur[u]; ~i; i = edge[i].nxt) {
        cur[u] = i;
        int v = edge[i].v, val = edge[i].val;
        if (val > 0 && dis[v] == dis[u] + 1) {
            int f = dfs(v, t, min(maxflow, val));
            if (f > 0) {
                edge[i].val -= f;
                edge[i ^ 1].val += f;
                return f;
            }
        }
    }
    return 0;
}
int dinic(int s, int t) {
    int flow = 0;
    while (bfs(s, t)) {
        while (int f = dfs(s, t, inf)) flow += f;
    }
    return flow;
}

int low[209][29], high[209][29], du[309];
char op[5];
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        memset(head, -1, sizeof(head));
        memset(low, 0, sizeof(low));
        memset(high, inf, sizeof(high));
        memset(du, 0, sizeof(du));
        gtot = 0;
        scanf("%d%d", &n, &m);
        int s = 0, t = n + m + 1, ss = n + m + 2, tt = n + m + 3;
        all = tt;
        int s1 = 0, s2 = 0;
        for (int i = 1; i <= n; i++) {
            int x;
            scanf("%d", &x);
            addedge(s, i, 0, -1);
            du[s] -= x;
            du[i] += x;
            s1 += x;
        }
        for (int i = 1; i <= m; i++) {
            int x;
            scanf("%d", &x);
            addedge(n + i, t, 0, -1);
            du[t] += x;
            du[n + i] -= x;
            s2 += x;
        }

        int c;
        scanf("%d", &c);
        int flag = 1;
        while (c--) {
            int u, v, x;
            scanf("%d %d %s %d", &u, &v, op, &x);
            if (u == 0 && v == 0) {
                for (int i = 1; i <= n; i++) {
                    for (int j = 1; j <= m; j++) {
                        if (op[0] == '>')
                            low[i][j] = max(low[i][j], x + 1);
                        else if (op[0] == '<')
                            high[i][j] = min(high[i][j], x - 1);
                        else if (op[0] == '=') {
                            low[i][j] = max(low[i][j], x), high[i][j] = min(high[i][j], x);
                            if (low[i][j] != x || high[i][j] != x) flag = 0;
                        }
                    }
                }
            } else if (u == 0) {
                for (int i = 1; i <= n; i++) {
                    if (op[0] == '>')
                        low[i][v] = max(low[i][v], x + 1);
                    else if (op[0] == '<')
                        high[i][v] = min(high[i][v], x - 1);
                    else if (op[0] == '=') {
                        low[i][v] = max(low[i][v], x), high[i][v] = min(high[i][v], x);
                        if (low[i][v] != x || high[i][v] != x) flag = 0;
                    }
                }
            } else if (v == 0) {
                for (int i = 1; i <= m; i++) {
                    if (op[0] == '>')
                        low[u][i] = max(low[u][i], x + 1);
                    else if (op[0] == '<')
                        high[u][i] = min(high[u][i], x - 1);
                    else {
                        low[u][i] = max(low[u][i], x), high[u][i] = min(high[u][i], x);
                        if (low[u][i] != x || high[u][i] != x) flag = 0;
                    }
                }
            } else {
                if (op[0] == '>')
                    low[u][v] = max(low[u][v], x + 1);
                else if (op[0] == '<')
                    high[u][v] = min(high[u][v], x - 1);
                else {
                    low[u][v] = max(low[u][v], x), high[u][v] = min(high[u][v], x);
                    if (low[u][v] != x || high[u][v] != x) flag = 0;
                }
            }
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                du[i] -= low[i][j];
                du[j + n] += low[i][j];
                addedge(i, n + j, high[i][j] - low[i][j], 1);
                if (high[i][j] < low[i][j]) flag = 0;
            }
        }

        int sum = 0;
        for (int i = s; i <= t; i++) {
            if (du[i] > 0) addedge(ss, i, du[i], -1), sum += du[i];
            if (du[i] < 0) addedge(i, tt, -du[i], -1);
        }
        if (s1 != s2 || !flag) {
            puts("IMPOSSIBLE");
            if (T) puts("");
            continue;
        }
        int f = dinic(ss, tt);

        if (f + s1 == sum) {
            for (int i = n + 1; i <= n + m; i++) {
                for (int j = head[i]; ~j; j = edge[j].nxt) {
                    int v = edge[j].v, val = edge[j].val;
                    low[v][i - n] += val;
                }
            }
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    if (j < m)
                        printf("%d ", low[i][j]);
                    else
                        printf("%d\n", low[i][j]);
                }
            }
        } else
            puts("IMPOSSIBLE");
        if (T) puts("");
    }
    return 0;
}
/*
2
2 3
8 10
5 6 7
4
0 2 > 2
2 1 = 3
2 3 > 2
2 3 < 5
2 2
4 5
6 7
1
1 1 > 10
*/