hdoj 3666 THE MATRIX PROBLEM 【差分约束】



THE MATRIX PROBLEM

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7410    Accepted Submission(s): 1896

Problem Description

You have been given a matrix C_N*M, each element E of C_N*M is positive and no more than 1000, The problem is that if there exist N numbers a1, a2, … an and M numbers b1, b2, …, bm, which satisfies that each elements in row-i multiplied with ai and each elements in column-j divided by bj, after this operation every element in this matrix is between L and U, L indicates the lowerbound and U indicates the upperbound of these elements.

 

Input

There are several test cases. You should process to the end of file.
Each case includes two parts, in part 1, there are four integers in one line, N,M,L,U, indicating the matrix has N rows and M columns, L is the lowerbound and U is the upperbound (1<=N、M<=400,1<=L<=U<=10000). In part 2, there are N lines, each line includes M integers, and they are the elements of the matrix.

 

Output

If there is a solution print "YES", else print "NO".

 

Sample Input

3 3 1 62 3 48 2 65 2 9

 

Sample Output

YES 

题意：给出一个N*M矩阵，下界L以及上界U，问你存不存在序列a[1...N] 和 b[1...M]，使得对矩阵里面的任一元素都有L <= Map[ i ][ j ] * a[ i ] / b[ j ] <= U 。存在输出YES，否则输出NO。

差分约束算法引子给出的是一系列未知数X1，X2，...Xn的不等式，让你求每个Xi(1 <= i <= n)的值。

和这道题很像吧？

思路：对于L <= Map[ i ][ j ] * a[ i ] / b[ j ] <= U 可转化为 log( L / Map[ i ][ j ] ) <= log( a[ i ] ) - log( b[ j ] ) <= log( U / Map[ i ][ j ]) 。log( a[ i ] ) 虚拟成 i 节点，log( b[ j ] ) 可以虚拟为 N + j节点。

上面条件的建图就不说了吧。。。

最后添加一个虚拟源点0，连通所有的点 ，给定一个权值0，因为点0是虚拟的，所以它和其它点间的关系不一定非要成立。 接下来就是最短路了。

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <queue>#include <algorithm>#define MAXN 1000#define MAXM 400000+10#define INF 1000000using namespace std;struct Edge{int from, to;double val;int next;}edge[MAXM];double dist[MAXN];int head[MAXN], top;int used[MAXN];bool vis[MAXN];double Map[410][410];int N, M;double L, U;void init(){top = 0;memset(head, -1, sizeof(head));}void addEdge(int u, int v, double w){    edge[top].from = u;    edge[top].to = v;    edge[top].val = w;    edge[top].next = head[u];head[u] = top++;}void getMap(){for(int i = 1; i <= N; i++){for(int j = 1; j <= M; j++){scanf("%lf", &Map[i][j]);addEdge(N+j, i, log(U/Map[i][j]));addEdge(i, N+j, -log(L/Map[i][j]));}}for(int i = 1; i <= N+M; i++)//建立源点 0  加入差分约束系统addEdge(0, i, 0);}void SPFA(){//queue<int> Q;int S[MAXN]; int top = 0;for(int i = 0; i <= N+M; i++){dist[i] = i==0 ? 0 : INF;used[i] = i==0 ? 1 : 0;vis[i] = i==0 ? true : false;}S[top++] = 0;while(top){int u = S[--top];vis[u] = false;for(int i = head[u]; i != -1; i = edge[i].next){Edge E = edge[i];if(dist[E.to] > dist[u] + E.val){dist[E.to] = dist[u] + E.val;if(!vis[E.to]){vis[E.to] = true;used[E.to]++;if(used[E.to] > sqrt(N+M)){printf("NO\n");return ;}                    S[top++] = E.to;}}}}printf("YES\n");}int main(){while(scanf("%d%d%lf%lf", &N, &M, &L, &U) != EOF){init();getMap();SPFA();}return 0;}