HDU 3666 THE MATRIX PROBLEM (差分约束判负环)

分类： ACM_HDU ACM_图论2013-10-27
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差分约束

题意:

给你一个N*M的矩阵C，问是否存在一个长度为N的数列a和长度为M的数列b使得所有C(i,j)*a(i)/b(j)在L , U范围内。

解题思路:

有两个式子 C(i,j)*a(i)/b(j) >= L , C(i,j)*a(i)/b(j) <= U ，两边均取对数得到 log(i) - log(j) <= log( U/C(i , j) ) 和 log(j) - log(i) <= log( C(i,j) / L) ，很显然的差分约束。。可惜用普通的队列版SPFA是会TLE的！网上说只需要入队sqrt(n)次就可以判断有负环了，不过没有具体的证明解释呀！所以正解应该是DFS版的也就是用栈的SPFA，因为要找负环，DFS找到一个负环后接下来相当于一直会绕着这个负环走个n+1次就跳出了，比BFS版判负环高效的多！

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/* **********************************************

Author : JayYe

Created Time: 2013-10-23 16:18:20

File Name : JayYe.cpp

*********************************************** */

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <algorithm>

using namespace std;

const int maxn = 800 + 5;

const int INF = 2000000000;

struct Edge{

int to, next;

double w;

}edge[maxn*maxn*2];

int head[maxn], T[maxn], E, q[maxn*maxn];

bool vis[maxn];

void init(int n) {

for(int i = 0;i <= n; i++)

head[i] = -1;

E = 0;

}

void newedge(int u, int to, double w) {

edge[E].to = to;

edge[E].w = w;

edge[E].next = head[u];

head[u] = E++;

}

double dis[maxn];

bool SPFA(int n) {

for(int i = 0;i <= n; i++) {

dis[i] = INF; vis[i] = T[i] = 0;

}

T[0] = 1;

dis[0] = 0;

vis[0] = 1;

int st = 0, ed = 0;

q[++ed] = 0;

while(ed) {

int u = q[ed--];

//if(st > n*n) return false;

vis[u] = 0;

for(int i = head[u];i != -1;i = edge[i].next) {

int to = edge[i].to;

double w = edge[i].w;

if(dis[to] > dis[u] + w) {

dis[to] = dis[u] + w;

if(!vis[to]) {

vis[to] = 1;

T[to]++;

if(T[to] > n + 1 ) return true;

q[++ed] = to;

}

}

}

}

return false;

}

int main() {

int n, m, L, U, c;

while(scanf("%d%d%d%d", &n, &m, &L, &U) != -1) {

init(n+m);

for(int i = 1;i <= n; i++) {

for(int j = 1;j <= m; j++) {

scanf("%d", &c);

newedge(j + n, i, log((double)U) - log((double)c));

newedge(i, j + n, log((double)c) - log((double)L));

}

}

for(int i = 1;i <= n+m; i++) newedge(0, i, 0);

if(!SPFA(n + m)) puts("YES");

else puts("NO");

}

return 0;

}