Budget
| Time Limit: 3000MS | Memory Limit: 65536K | |||
| Total Submissions: 5621 | Accepted: 2140 | Special Judge | ||
Description
We are supposed to make a budget proposal for this multi-site competition. The budget proposal is a matrix where the rows represent different kinds of expenses and the columns represent different sites. We had a meeting about this,
some time ago where we discussed the sums over different kinds of expenses and sums over different sites. There was also some talk about special constraints: someone mentioned that Computer Center would need at least 2000K Rials for food and someone from Sharif
Authorities argued they wouldn't use more than 30000K Rials for T-shirts. Anyway, we are sure there was more; we will go and try to find some notes from that meeting.
And, by the way, no one really reads budget proposals anyway, so we'll just have to make sure that it sums up properly and meets all constraints.
And, by the way, no one really reads budget proposals anyway, so we'll just have to make sure that it sums up properly and meets all constraints.
Input
The first line of the input contains an integer N, giving the number of test cases. The next line is empty, then, test cases follow: The first line of each test case contains two integers, m and n, giving the number of rows and
columns (m <= 200, n <= 20). The second line contains m integers, giving the row sums of the matrix. The third line contains n integers, giving the column sums of the matrix. The fourth line contains an integer c (c < 1000) giving the number of constraints.
The next c lines contain the constraints. There is an empty line after each test case.
Each constraint consists of two integers r and q, specifying some entry (or entries) in the matrix (the upper left corner is 1 1 and 0 is interpreted as "ALL", i.e. 4 0 means all entries on the fourth row and 0 0 means the entire matrix), one element from the set {<, =, >} and one integer v, with the obvious interpretation. For instance, the constraint 1 2 > 5 means that the cell in the 1st row and 2nd column must have an entry strictly greater than 5, and the constraint 4 0 = 3 means that all elements in the fourth row should be equal to 3.
Each constraint consists of two integers r and q, specifying some entry (or entries) in the matrix (the upper left corner is 1 1 and 0 is interpreted as "ALL", i.e. 4 0 means all entries on the fourth row and 0 0 means the entire matrix), one element from the set {<, =, >} and one integer v, with the obvious interpretation. For instance, the constraint 1 2 > 5 means that the cell in the 1st row and 2nd column must have an entry strictly greater than 5, and the constraint 4 0 = 3 means that all elements in the fourth row should be equal to 3.
Output
For each case output a matrix of non-negative integers meeting the above constraints or the string "IMPOSSIBLE" if no legal solution exists. Put
one empty line between matrices.
Sample Input
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
Sample Output
2 3 3 3 3 4 IMPOSSIBLE
題意：有一個矩陣，n行m列，給出每行元素的和、每列元素的和，並且給出c個約束，每個約束(u, v) = a代表元素(u, v)必須等於a（>或<時以此類推），若u=0，則代表第v列元素必須全部=（或>、<）a。問是否存在一個滿足所有約束條件的矩陣，若存在，輸出這個矩陣。
思路：設源點s，匯點t，每行和每列看成是一個結點，s到行結點i連一條[sum[i], sum[i]]的邊，列節點i到t連一條[sum[i], sum[i]]的邊，然後行結點到列節點連滿足上下界的邊，再虛設一個超級源點ss和超級匯點tt，求一次可行流就可以了。
AC代碼：
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#include <ctime>
#include <vector>
#include <algorithm>
#define ll long long
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
using namespace std;
const int INF = 1e9;
const int maxn = 505;
struct Edge
{
int u, v, cap, flow, next;
} et[maxn * maxn];
int low[maxn], cnt[maxn], dis[maxn], pre[maxn], cur[maxn], eh[maxn], du[maxn];
int L[maxn][maxn], R[maxn][maxn];
int n, m, s, t, num, ss, tt, sum1, sum2, sum;
void init()
{
memset(eh, -1, sizeof(eh));
memset(du, 0, sizeof(du));
for(int i = 0; i <= n; i++)
for(int j = 0; j <= m; j++)
{
L[i][j] = 0;
R[i][j] = INF;
}
num = sum1 = sum2 = sum = 0;
}
void add(int u, int v, int cap, int flow)
{
Edge e = {u, v, cap, flow, eh[u]};
et[num] = e;
eh[u] = num++;
}
void addedge(int u, int v, int cap)
{
add(u, v, cap, 0);
add(v, u, 0, 0);
}
int isap(int s, int t, int nv)
{
int u, v, now, flow = 0;
memset(low, 0, sizeof(low));
memset(cnt, 0, sizeof(cnt));
memset(dis, 0, sizeof(dis));
for(u = 0; u <= nv; u++) cur[u] = eh[u];
low[s] = INF, cnt[0] = nv, u = s;
while(dis[s] < nv)
{
for(now = cur[u]; now != -1; now = et[now].next)
if(et[now].cap - et[now].flow && dis[u] == dis[v = et[now].v] + 1) break;
if(now != -1)
{
cur[u] = pre[v] = now;
low[v] = min(low[u], et[now].cap - et[now].flow);
u = v;
if(u == t)
{
for(; u != s; u = et[pre[u]].u)
{
et[pre[u]].flow += low[t];
et[pre[u]^1].flow -= low[t];
}
flow += low[t];
low[s] = INF;
}
}
else
{
if(--cnt[dis[u]] == 0) break;
dis[u] = nv, cur[u] = eh[u];
for(now = eh[u]; now != -1; now = et[now].next)
if(et[now].cap - et[now].flow && dis[u] > dis[et[now].v] + 1)
dis[u] = dis[et[now].v] + 1;
cnt[dis[u]]++;
if(u != s) u = et[pre[u]].u;
}
}
return flow;
}
bool judge()
{
if(sum1 != sum2) return false;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
if(L[i][j] > R[i][j]) return false;
addedge(i, j + n, R[i][j] - L[i][j]);
du[i] -= L[i][j];
du[j + n] += L[i][j];
}
return true;
}
void solve()
{
ss = t + 1, tt = t + 2;
sum = 0;
for(int i = 0; i <= t; i++)
{
if(du[i] > 0)
{
addedge(ss, i, du[i]);
sum += du[i];
}
else if(du[i] < 0) addedge(i, tt, -du[i]);
}
addedge(t, s, INF);
int tmp = isap(ss, tt, tt + 1);
if(tmp != sum)
{
printf("IMPOSSIBLE\n");
return;
}
int cc = 0;
for(int i = 1; i <= n; i++)
{
for(int j = 1; j < m; j++)
{
printf("%d ", et[cc].flow + L[i][j]);
cc += 2;
}
printf("%d\n", et[cc].flow + L[i][m]);
cc += 2;
}
}
int main()
{
int ca, a, c, nn, mm;
char ch[5];
scanf("%d", &ca);
while(ca--)
{
scanf("%d%d", &n, &m);
init();
s = 0, t = n + m + 1;
for(int i = 1; i <= n; i++)
{
scanf("%d", &c); //行結點[c, c]
du[s] -= c;
du[i] += c;
sum1 += c;
}
for(int i = n + 1; i <= n + m; i++)
{
scanf("%d", &c); //列節點[c, c]
du[i] -= c;
du[t] += c;
sum2 += c;
}
scanf("%d", &c);
while(c--)
{
scanf("%d%d%s%d", &nn, &mm, ch, &a);
int r1 = nn, r2 = nn, c1 = mm, c2 = mm;
if(nn == 0) r1 = 1, r2 = n;
if(mm == 0) c1 = 1, c2 = m;
for(int i = r1; i <= r2; i++)
for(int j = c1; j <= c2; j++)
{
if(ch[0] == '=')
{
L[i][j] = max(L[i][j], a);
R[i][j] = min(R[i][j], a);
}
else if(ch[0] == '>') L[i][j] = max(L[i][j], a + 1);
else R[i][j] = min(R[i][j], a - 1);
}
}
if(judge()) solve();
else printf("IMPOSSIBLE\n");
puts("");
}
return 0;
}