HDU_3666_THE MATRIX PROBLEM(差分约束+spfa+slf优化)

THE MATRIX PROBLEM

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7644    Accepted Submission(s): 1973

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 

 

Source

2010 Asia Regional Harbin

 

题意：给你一个 N*M 的矩阵 x[N][M]，问是否存在这样的两个序列( a[1], a[2], a[3], …… a[N] )，( b[1], b[2], b[3], …… b[M] )，使得对于任意的 x[i][j]，都有 L <= x[i][j] * a[i] / b[j] <= U。

分析；差分约束问题。

差分约束：即对于给定的不等式 a - b <= c，对应于最短路中的 d[v] <= d[u] + dist[u][v] 这条性质，可以连边 b -> a，边权为 c；这样对于多组不等式连边构图之后，得到源点到每个点的最短路即为不等式的最小解。而如果图中存在负圈，那么无解。

此题也可以演变为差分约束问题，对不等式进行变形可得：L / x[i][j] <= a[i] / b[j] <= U / x[i][j]；如何把它变成相减的形式呢？可以两边同时取对数，即可得：log(a[i]) - log(b[j]) <= log(U / x[i][j]), log(b[j]) - log(a[i]) <= -log(L / x[i][j])；那么可以得到两组边( j + N, i, log(U / x[i][j]) )，( i, j + N, -log(L / x[i][j]) )； 图建好后，跑最短路判断即可。但是此题就在此会出问题了，spfa判断负圈，是单点入队次数超过 N+M 次即可判定为存在负圈，如果用最原始的spfa会超时，这里用SLF优化后即可过。SLF：设当前点为 v，队首元素为u，若dist[v] < dist[u]，即把 v 加入到队列的头部，否则加入到队列尾部。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3666

代码清单：

#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <ctime>#include <vector>#include <cctype>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define end() return 0typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;const int maxn = 400 + 5;const double maxd = 1000000000.0 + 5.0;struct Edge{    int to;    int next;    double dis;};int N, M, L, U;int x[maxn][maxn];int num;int head[maxn * 2];Edge graph[maxn * maxn * 2];double d[maxn * 2];int cnt[maxn * 2];bool vis[maxn * 2];void init(){    memset(graph, 0, sizeof(graph));    memset(head, -1, sizeof(head));    num = 0;}void addEdge(int u, int v, double dis){    graph[num].to = v;    graph[num].dis = dis;    graph[num].next = head[u];    head[u] = num++;}void input(){    for(int i = 1; i <= N; i++){        for(int j = 1; j <= M; j++){            scanf("%d", &x[i][j]);        }    }}void createGraph(){    for(int i = 1; i <= M + N; i++) addEdge(0, i, 0.0);    for(int i = 1; i <= N; i++){        for(int j = 1; j <= M; j++){            addEdge(N + j, i, log(1.0 * U / x[i][j]));            addEdge(i, N + j, -log(1.0 * L / x[i][j]));        }    }}bool spfa(){    fill(d, d + 1 + M + N, maxd);    memset(vis, false, sizeof(vis));    memset(cnt, 0, sizeof(cnt));    d[0] = 0.0;    cnt[0] = 1;    vis[0] = true;    deque <int> q;    q.push_front(0);    while(!q.empty()){        int u = q.front(); q.pop_front();        vis[u] = false;        for(int i = head[u]; i != -1; i = graph[i].next){            int v = graph[i].to;            double dis = graph[i].dis;            if(d[v] > d[u] + dis){                d[v] = d[u] + dis;                if(!vis[v]){                    vis[v] = true;                    cnt[v]++;                    if(cnt[v] >= N + M) return false;                    if(!q.empty()){                        if(d[v] < d[q.front()]) q.push_front(v);                        else q.push_back(v);                    }                    else q.push_front(v);                }            }        }    }return true;}void solve(){    createGraph();    if(spfa()) puts("YES");    else puts("NO");}int main(){    while(scanf("%d%d%d%d", &N, &M, &L, &U) != EOF){        init();        input();        solve();    }   end();}