POJ 2396 Budget

Budget

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2396
64-bit integer IO format: %lld      Java class name: Main

 

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.

 

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.

 

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 

Source

Tehran 2003 Preliminary

 

解题：贫道正在赶。。。这题建图正尼玛。。。。

 

绝世好。。。啊。。

 

有源汇的最大流，把原来的汇点到源点连一条无穷的边，然后根据无源汇求建图的方法，建立超级源汇点，满流，则最大流就是此有源汇的上下界最大流。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 500;
int d[maxn],low[maxn][maxn],high[maxn][maxn];
int tot,row,col,colsum[maxn],rowsum[maxn],S,T;
int ans[maxn][maxn],flow[maxn][maxn],de[maxn];
bool bfs() {
    memset(d,-1,sizeof(d));
    queue<int>q;
    d[S] = 1;
    q.push(S);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = 0; i <= T; ++i)
            if(flow[u][i] && d[i] == -1) {
                d[i] = d[u] + 1;
                q.push(i);
            }
    }
    return d[T] > -1;
}
int dfs(int u,int low) {
    if(u == T) return low;
    int tmp = 0,a;
    for(int i = 0; i <= T; ++i) {
        if(flow[u][i] && d[i] == d[u] + 1&&(a=dfs(i,min(low,flow[u][i])))) {
            flow[u][i] -= a;
            flow[i][u] += a;
            low -= a;
            tmp += a;
            if(!low) break;
        }
    }
    if(!tmp) d[u] = -1;
    return tmp;
}
int dinic(){
    int ans = 0;
    while(bfs()) ans += dfs(S,INF);
    return ans;
}
int main() {
    int cs,con;
    bool cao = false;
    scanf("%d",&cs);
    while(cs--) {
        if(cao) puts("");
        cao = true;
        scanf("%d %d",&row,&col);
        for(int i = 1; i <= row; ++i) scanf("%d",rowsum+i);
        for(int i = 1; i <= col; ++i) scanf("%d",colsum+i);
        memset(low,0,sizeof(low));
        memset(flow,0,sizeof(flow));
        memset(de,0,sizeof(de));
        for(int i = 1; i <= row; ++i)
            for(int j = 1; j <= col; ++j)
                high[i][row + j] = INF;
        scanf("%d",&con);
        while(con--) {
            int u,v,w;
            char op[5];
            scanf("%d %d %s %d",&u,&v,op,&w);
            if(!u && v) {
                if(op[0] == '=') {
                    for(int i = 1; i <= row; ++i)
                        high[i][v+row] = low[i][v+row] = w;
                } else if(op[0] == '>') {
                    for(int i = 1; i <= row; ++i)
                        low[i][v+row] = max(low[i][v+row],w+1);
                } else if(op[0] == '<') {
                    for(int i = 1; i <= row; ++i)
                        high[i][v+row] = min(high[i][v+row],w-1);
                }
            } else if(u && !v) {
                if(op[0] == '=') {
                    for(int i = 1; i <= col; ++i)
                        low[u][i+row] = high[u][i+row] = w;
                } else if(op[0] == '>') {
                    for(int i = 1; i <= col; ++i)
                        low[u][i+row] = max(low[u][i+row],w+1);
                } else if(op[0] == '<') {
                    for(int i = 1; i <= col; ++i)
                        high[u][i+row] = min(high[u][i+row],w-1);
                }
            } else if(!u && !v) {
                if(op[0] == '=') {
                    for(int i = 1; i <= row; ++i)
                        for(int j = 1; j <= col; ++j)
                            low[i][j+row] = high[i][j+row] = w;
                } else if(op[0] == '>') {
                    for(int i = 1; i <= row; ++i)
                        for(int j = 1; j <= col; ++j)
                            low[i][j+row] = max(low[i][j+row],w+1);
                } else if(op[0] == '<') {
                    for(int i = 1; i <= row; ++i)
                        for(int j = 1; j <= col; ++j)
                            high[i][j+row] = min(high[i][j+row],w-1);
                }
            } else if(u && v) {
                if(op[0] == '=') low[u][v+row] = high[u][v+row] = w;
                else if(op[0] == '>') low[u][v+row] = max(low[u][v+row],w+1);
                else if(op[0] == '<') high[u][v+row] = min(high[u][v+row],w-1);
            }
        }
        S = 0;
        T = row + col + 1;
        for(int i = 1; i <= row; ++i)
            low[S][i] = high[S][i] = rowsum[i];
        for(int i = 1; i <= col; ++i)
            low[i+row][T] = high[i+row][T] = colsum[i];
        int SS = T+1,TT = T + 2,sum = 0;
        for(int i = 1; i <= row; ++i)
            for(int j = 1; j <= col; ++j){
                flow[i][j+row] = high[i][j+row] - low[i][j+row];
                de[i] -= low[i][j+row];
                de[j+row] += low[i][j+row];
            }
        for(int i = 1; i <= row; ++i){
            flow[S][i] = high[S][i] - low[S][i];
            de[S] -= low[S][i];
            de[i] += low[S][i];
        }
        for(int j = 1; j <= col; ++j){
            flow[j+row][T] = high[j+row][T] - low[j+row][T];
            de[j+row] -= low[j+row][T];
            de[T] += low[j+row][T];
        }
        for(int i = 0; i <= T; ++i){
            if(de[i] > 0) {sum += de[i];flow[SS][i] = de[i];}
            if(de[i] < 0) flow[i][TT] = -de[i];
        }
        flow[T][S] = INF;
        S = SS;
        T = TT;
        if(dinic() == sum){
            S = 0;
            T = row + col + 1;
            dinic();
            for(int i = 1; i <= row; ++i)
                for(int j = 1; j <= col; ++j)
                    printf("%d%c",flow[j+row][i]+low[i][row+j],j == col?'
':' ');
        }else puts("IMPOSSIBLE");
    }
    return 0;
}