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

#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 = 345,T = 1234;
struct arc{
    int u,v,w;
    arc(int x = 0,int y = 0,int z = 0){
        u = x;
        v = y;
        w = z;
    }
};
vector<int>g[maxn];
vector<int>G[maxn];
vector<arc>e;
char mp[maxn][maxn];
int n,color[maxn],d[maxn],cnt[maxn];
bool in[maxn];
void dfs(int u,int c){
    color[u] = c;
    for(int i = 0; i < g[u].size(); i++)
        if(!color[g[u][i]]) dfs(g[u][i],-c);
}
void add(int u,int v,int w){
    e.push_back(arc(u,v,w));
    G[u].push_back(e.size()-1);
}
bool spfa(){
    queue<int>q;
    for(int i = 0; i < n; i++){
        d[i] = 0;
        cnt[i] = 1;
        in[i] = true;
        q.push(i);
    }
    while(!q.empty()){
        int u = q.front();
        q.pop();
        in[u] = false;
        for(int i = 0; i < G[u].size(); i++){
            arc &temp = e[G[u][i]];
            if(d[temp.v] > d[u]+temp.w){
                d[temp.v] = d[u]+temp.w;
                if(!in[temp.v]){
                    in[temp.v] = true;
                    cnt[temp.v]++;
                    if(cnt[temp.v] > n) return true;
                    q.push(temp.v);
                }
            }
        }
    }
    return false;
}
bool solve(){
    for(int i = 0; i < n; i++) if(!color[i]) dfs(i,1);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < g[i].size(); j++)
            if(color[i] == color[g[i][j]]) return false;
    for(int i = 0; i < n; i++){
        for(int j = i+1; j < n; j++){
            if(mp[i][j] == '1')
                color[i] == 1?add(i,j,-T):add(j,i,-T);
            else color[i] == 1?add(j,i,T-1):add(i,j,T-1);
        }
    }
    return !spfa();
}
int main() {
    int ks,i,j;
    scanf("%d",&ks);
    while(ks--){
        scanf("%d",&n);
        for(i = 0; i <= n; i++){
            g[i].clear();
            G[i].clear();
            color[i] = 0;
        }
        e.clear();
        for(i = 0; i < n; i++){
            scanf("%s",mp[i]);
            for(j = 0; j < n; j++)
                if(mp[i][j] == '1') g[i].push_back(j);
        }
        solve()?puts("Yes"):puts("No");
    }
    return 0;
}