[luogu2474 SCOI2008] Balance (Floyd differential constraint)

Transmission Door

Solution

Due to the weight of only three cases, the thought of using differential constraints.
Due to the small range, the thought can be floyed to seek differential constraints, the balance of violence to seek the other side

Code

#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include < Algorithm> #define F (I,A,B) for (register int i= (a), i<= (b); i++) #define E (I,u) for (register int i=head[u],v;i;i=e[i    ].NXT) using namespace Std;inline int read () {int X=0,f=1;char c=getchar ();    while (!isdigit (c)) {if (c== '-') F=-f;c=getchar ();}    while (IsDigit (c)) x= (x<<1) + (x<<3) +c-48,c=getchar (); return x*f;}    const int N=60;int n,a,b,ans1,ans2,ans3;int Mx[n][n],mi[n][n];char ch[n];int Main () {N=read (), A=read (), B=read ();        F (i,1,n) {scanf ("%s", ch+1);            F (J,1,n) if (ch[j]== ' = ') mx[i][j]=mi[i][j]=0;            else if (ch[j]== '-') mx[i][j]=-1,mi[i][j]=-2;            else if (ch[j]== ' + ') mx[i][j]=2,mi[i][j]=1;    else mx[i][j]=2,mi[i][j]=-2; } f (K,1,n) f (i,1,n) if (i!=k) f (j,1,n) if (i!=j&&k!=j) mx[i][j]=min (Mx[i][j],mx[i][k]+mx[k][j]), MI I    [J]=max (Mi[i][j],mi[i][k]+mi[k][j]); F (i,1,n) if (i!=a&&i!=b) F(j,i+1,n) if (j!=a&&j!=b) {if (mi[a][i]>mx[j][b]| |        MI[A][J]&GT;MX[I][B]) ans1++; if (mx[a][i]<mi[j][b]| |        MX[A][J]&LT;MI[I][B]) ans3++; if ((mi[a][i]==mx[a][i]&&mi[b][j]==mx[b][j]&&mi[a][i]==mi[j][b)) | | (Mi[a][j]==mx[a][j]&&mi[b][i]==mx[b][i]&&mi[a][j]==mi[i][b]))    ans2++;    } printf ("%d%d%d", ANS1,ANS2,ANS3); return 0;}

[luogu2474 SCOI2008] Balance (Floyd differential constraint)