You're given a matrix A of size n × n.
Let's call the matrix with nonnegative elements magic if it is symmetric (so aij = aji), aii = 0 and aij ≤ max(aik, ajk) for all triples i, j, k. Note that i, j, k do not need to be distinct.
Determine if the matrix is magic.
As the input/output can reach very huge size it is recommended to use fast input/output methods: for example, prefer to usescanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java.
The first line contains integer n (1 ≤ n ≤ 2500) — the size of the matrix A.
Each of the next n lines contains n integers aij (0 ≤ aij < 109) — the elements of the matrix A.
Note that the given matrix not necessarily is symmetric and can be arbitrary.
Print ''MAGIC" (without quotes) if the given matrix A is magic. Otherwise print ''NOT MAGIC".
3 0 1 2 1 0 2 2 2 0
MAGIC
2 0 1 2 3
NOT MAGIC
4 0 1 2 3 1 0 3 4 2 3 0 5 3 4 5 0
NOT MAGIC
题意：给你一个n*n的矩阵，让你判断这个矩阵是不是魔力矩阵，魔力矩阵的定义为：1.对角线都为0. 2.左下角的数和右上角的数对称相等. 3.对于任意一个格子(i,j)要满足对于任意的k，a[i][j]<=max(a[i][k],a[k][j]),k为1~n中的任意数，可以与i，j相等。
思路：有两种思路，第一种一种比较容易想，因为要满足对于任意的k，a[i][j]<=max(a[i][k],a[k][j])，k为任意数，那么a[i][j]就满足a[i][j]<=max(a[i][k],a[j][k]),因为满足前两种条件的前提下a[k][j]=a[j][k].那么再把不等式转换，即变成a[i][j]要小于等于n对i,j行上下对应的两个数的最大值的最小值，因为k是任意取的.那么我们可以先把所有的点的x坐标，y坐标，大小放入结构体中，然后根据大小从小到大排序.然后开一个bitset<maxn>bt[maxn],b[x]表示的是x行中比a[i][j]小的列数的表示（如果x行当前的列数小于a[i][j]，该位就置为1），那么对于现在这个数，所有小于这个数的都在i,j行的bitset里，如果这两行的bitset交非空,说明存在某个k，使a[i][j]>a[i][k]且a[i][j]>a[j][k],这样就是不符合条件的.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 2505
int a[maxn][maxn];
struct node{
int len,l,r;
}e[maxn*maxn/2];
bool cmp(node a,node b){
return a.len<b.len;
}
bitset<maxn>bt[maxn];
int main()
{
int n,m,i,j,flag;
while(scanf("%d",&n)!=EOF)
{
flag=1;
int tot=0;
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
scanf("%d",&a[i][j]);
if(i<j){
tot++;
e[tot].len=a[i][j];
e[tot].l=i;e[tot].r=j;
}
}
}
for(i=1;i<=n;i++){
if(a[i][i]!=0){
flag=0;break;
}
}
if(flag==0){
printf("NOT MAGIC
");continue;
}
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
if(a[i][j]!=a[j][i]){
flag=0;break;
}
}
if(!flag)break;
}
if(flag==0){
printf("NOT MAGIC
");continue;
}
sort(e+1,e+1+tot,cmp);
int t=1;
for(i=1;i<=tot;i++){
while(t<=tot && e[t].len<e[i].len ){
bt[e[t].l ][e[t].r ]=1;
bt[e[t].r ][e[t].l ]=1;
t++;
}
if((bt[e[i].l ]&bt[e[i].r ] ).any() ){
flag=0;break;
}
}
if(flag==0)printf("NOT MAGIC
");
else printf("MAGIC
");
}
return 0;
}
第二种思路：是把这个矩阵看做一张图，a[i][j]表示i和j点之间连一条a[i][j]的边，我们设b[i][j]为i节点到j节点之间所有路径最长边的最小值，那么根据定义可得a[i][j]>=b[i][j].然后如果是魔力矩阵，那么要满足a[i][j]<=max(a[i][k]+a[k][j]),因为a[i][k]<=max(a[i][k1]+a[k1][k])...可以多次递归下去，所以a[i][j]<=max(a[i][k1],a[k1][k2]+...+a[km][j),即相当于a[i][j]<=b[i][j],所以a[i][j]=b[i][j].接下来我们就要先的到b[i][j],这里我们可以用最小生成树做，因为最小生成树每次都是加最短的边，所以能够保证使得最大的边最小.把最小生成树求出来之后，我们枚举1~n的每一个点为根节点，dfs一遍所有点，记录根到其他所有点的最小生成树路径中的最小边就行了.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
#define inf 99999999
#define pi acos(-1.0)
#define maxn 2505
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef long double ldb;
int a[maxn][maxn];
struct node{
int len,l,r;
}e[maxn*maxn/2];
int pre[maxn],ran[maxn],num[maxn],maxx[maxn];
struct edg{
int next,to,len;
}edge[2*maxn];
int first[maxn];
int findset(int x){
int i,j=x,r=x;
while(r!=pre[r])r=pre[r];
while(j!=pre[j]){
i=pre[j];
pre[j]=r;
j=i;
}
return r;
}
bool cmp(node a,node b){
return a.len<b.len;
}
int flag;
void dfs(int u,int father,int x)
{
int i,j,v;
for(i=first[u];i!=-1;i=edge[i].next){
v=edge[i].to;
if(v==father)continue;
maxx[v]=max(maxx[u],edge[i].len);
if(a[x][v]!=maxx[v]){
flag=0;break;
}
dfs(v,u,x);
if(flag==0)break;
}
}
int main()
{
int n,m,i,j;
while(scanf("%d",&n)!=EOF)
{
flag=1;
int tot=0;
for(i=1;i<=n;i++){
pre[i]=i;ran[i]=0;num[i]=1;
for(j=1;j<=n;j++){
scanf("%d",&a[i][j]);
if(i<j){
tot++;
e[tot].len=a[i][j];
e[tot].l=i;e[tot].r=j;
}
}
}
for(i=1;i<=n;i++){
if(a[i][i]!=0){
flag=0;break;
}
}
if(flag==0){
printf("NOT MAGIC
");continue;
}
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
if(a[i][j]!=a[j][i]){
flag=0;break;
}
}
if(!flag)break;
}
if(flag==0){
printf("NOT MAGIC
");continue;
}
sort(e+1,e+1+tot,cmp);
int t1,t2,u,v,x,y;
int t=0;
memset(first,-1,sizeof(first));
for(i=1;i<=tot;i++){
u=e[i].l;v=e[i].r;
x=findset(u);
y=findset(v);
if(x==y)continue;
t++;
edge[t].next=first[u];edge[t].to=v;edge[t].len=a[u][v];
first[u]=t;
t++;
edge[t].next=first[v];edge[t].to=u;edge[t].len=a[u][v];
first[v]=t;
if(ran[x]>ran[y]){
pre[y]=x;
num[x]+=num[y];
if(num[x]==n)break;
}
else{
pre[x]=y;
num[y]+=num[x];
if(num[y]==n)break;
if(ran[x]==ran[y])ran[y]++;
}
}
for(j=1;j<=n;j++){
maxx[j]=0;
dfs(j,0,j);
if(flag==0)break;
}
if(flag)printf("MAGIC
");
else printf("NOT MAGIC
");
}
return 0;
}