HDU 4598 Difference

Difference

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 581    Accepted Submission(s): 152

Problem Description

A graph is a difference if every vertex vi can be assigned a real number ai and there exists a positive real number T such that
(a) |a _i| < T for all i and 
(b) (v _i, v _j) in E <=> |a _i - a _j| >= T,
where E is the set of the edges. 
Now given a graph, please recognize it whether it is a difference.

 

 

Input

The first line of input contains one integer TC(1<=TC<=25), the number of test cases.
Then TC test cases follow. For each test case, the first line contains one integer N(1<=N<=300), the number of vertexes in the graph. Then N lines follow, each of the N line contains a string of length N. The j-th character in the i-th line is "1" if (v _i, v _j) in E, and it is "0" otherwise. The i-th character in the i-th line will be always "0". It is guaranteed that the j-th character in the i-th line will be the same as the i-th character in the j-th line.

 

 

Output

For each test case, output a string in one line. Output "Yes" if the graph is a difference, and "No" if it is not a difference.

 

 

Sample Input

3

4

0011

0001

1000

1100

4

0111

1001

1001

1110

3

000

000

000

 

 

Sample Output

Yes

No

Yes

 

Hint

In sample 1, it can let T=3 and a[sub]1[/sub]=-2, a[sub]2[/sub]=-1, a[sub]3[/sub]=1, a[sub]4[/sub]=2.

 

 

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

 

 

Recommend

liuyiding

 

解题：差分约束，完全不懂为什么要这样子。学渣的无奈啊

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 #include <climits>

  7 #include <vector>

  8 #include <queue>

  9 #include <cstdlib>

 10 #include <string>

 11 #include <set>

 12 #include <stack>

 13 #define LL long long

 14 #define pii pair<int,int>

 15 #define INF 0x3f3f3f3f

 16 using namespace std;

 17 const int maxn = 345,T = 1234;

 18 struct arc{

 19     int u,v,w;

 20     arc(int x = 0,int y = 0,int z = 0){

 21         u = x;

 22         v = y;

 23         w = z;

 24     }

 25 };

 26 vector<int>g[maxn];

 27 vector<int>G[maxn];

 28 vector<arc>e;

 29 char mp[maxn][maxn];

 30 int n,color[maxn],d[maxn],cnt[maxn];

 31 bool in[maxn];

 32 void dfs(int u,int c){

 33     color[u] = c;

 34     for(int i = 0; i < g[u].size(); i++)

 35         if(!color[g[u][i]]) dfs(g[u][i],-c);

 36 }

 37 void add(int u,int v,int w){

 38     e.push_back(arc(u,v,w));

 39     G[u].push_back(e.size()-1);

 40 }

 41 bool spfa(){

 42     queue<int>q;

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

 44         d[i] = 0;

 45         cnt[i] = 1;

 46         in[i] = true;

 47         q.push(i);

 48     }

 49     while(!q.empty()){

 50         int u = q.front();

 51         q.pop();

 52         in[u] = false;

 53         for(int i = 0; i < G[u].size(); i++){

 54             arc &temp = e[G[u][i]];

 55             if(d[temp.v] > d[u]+temp.w){

 56                 d[temp.v] = d[u]+temp.w;

 57                 if(!in[temp.v]){

 58                     in[temp.v] = true;

 59                     cnt[temp.v]++;

 60                     if(cnt[temp.v] > n) return true;

 61                     q.push(temp.v);

 62                 }

 63             }

 64         }

 65     }

 66     return false;

 67 }

 68 bool solve(){

 69     for(int i = 0; i < n; i++) if(!color[i]) dfs(i,1);

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

 71         for(int j = 0; j < g[i].size(); j++)

 72             if(color[i] == color[g[i][j]]) return false;

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

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

 75             if(mp[i][j] == '1')

 76                 color[i] == 1?add(i,j,-T):add(j,i,-T);

 77             else color[i] == 1?add(j,i,T-1):add(i,j,T-1);

 78         }

 79     }

 80     return !spfa();

 81 }

 82 int main() {

 83     int ks,i,j;

 84     scanf("%d",&ks);

 85     while(ks--){

 86         scanf("%d",&n);

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

 88             g[i].clear();

 89             G[i].clear();

 90             color[i] = 0;

 91         }

 92         e.clear();

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

 94             scanf("%s",mp[i]);

 95             for(j = 0; j < n; j++)

 96                 if(mp[i][j] == '1') g[i].push_back(j);

 97         }

 98         solve()?puts("Yes"):puts("No");

 99     }

100     return 0;

101 }

View Code