bzoj1077: [SCOI2008]天平洛谷P2474

https://www.luogu.com.cn/problem/P2474

https://lydsy.com/JudgeOnline/problem.php?id=1077

優先 1<=i<=3,所以有初始條件j-i>=-2,j-i<=2

dismi[i][j]>=T,dismx[i][j]<=T

然後s[i][j]="+" -> j-i<=-1   s[i][j]='-'   ->  j-i>=1   s[i][j]='='  ->j-i>=0,j-i<=0

數據很小，用floyed跑最短路和最長路就行了。

c1->A+B>=C+D+1

c2->A+B>=C+D,A+B<=C+D

c3->A+B<=C+D-1

用dismi[]和dismx[]判斷一下就行了。

#include<bits/stdc++.h>
using namespace std;

const int maxl=55;

int n,cnt,ans,A,B,c1,c2,c3;
int dismi[maxl][maxl],dismx[maxl][maxl];
queue<int> q;
char s[maxl][maxl];

inline void prework()
{
	scanf("%d%d%d",&n,&A,&B);
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		if(i!=j)
			dismi[i][j]=-2,dismx[i][j]=2;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			if(s[i][j]=='+')
				dismi[i][j]=-2,dismx[i][j]=-1;
			if(s[i][j]=='-')
				dismi[i][j]=1,dismx[i][j]=2;
			if(s[i][j]=='=')
				dismi[i][j]=dismx[i][j]=0;		
		}

}

inline void mainwork()
{
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
			if(i!=j && j!=k && i!=k)
			{
				dismi[i][j]=max(dismi[i][j],dismi[i][k]+dismi[k][j]);
				dismx[i][j]=min(dismx[i][j],dismx[i][k]+dismx[k][j]);
			}
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
		if(i!=A && i!=B && j!=A && j!=B)
		{
			if(dismi[i][A]+dismi[j][B]>=1 || dismi[j][A]+dismi[i][B]>=1)
				c1++;
			if(dismx[i][A]+dismx[j][B]<=-1 || dismx[j][A]+dismx[i][B]<=-1)
				c3++;
			if((dismi[i][A]+dismi[j][B]>=0 && dismx[i][A]+dismx[j][B]<=0) ||
			   (dismi[j][A]+dismi[i][B]>=0 && dismx[j][A]+dismx[i][B]<=0)  )
				c2++;
		}
}

inline void print()
{
	printf("%d %d %d",c1,c2,c3);
}

int main()
{
	prework();
	mainwork();
	print();
	return 0;
}