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51nod 1340 差分约束

思路:

带未知量的Floyd

很强

http://yousiki.net/index.php/archives/87/

//By SiriusRen#include <bits/stdc++.h>using namespace std;const int B=55;int cases,n,m1,m2,xx,yy,zz;typedef long long ll;ll f[55][55][115],l,r,inf;void up(ll &x,ll y){x=min(x,y);}signed main(){    scanf("%d",&cases);    while(cases--){        memset(f,0x3f,sizeof(f));        scanf("%d%d%d",&n,&m1,&m2),l=n,r=inf=f[0][0][0];        for(int i=0;i<=n;i++)f[i][i][B]=0,i?f[i][i-1][B]=-1:0;        up(f[0][n][B+1],0);up(f[n][0][B-1],0);        for(int i=1;i<=m1;i++)scanf("%d%d%d",&xx,&yy,&zz),up(f[yy][xx][B+(xx>=yy)],-zz);        for(int i=1;i<=m2;i++)scanf("%d%d%d",&xx,&yy,&zz),up(f[xx][yy][B-(xx>=yy)],zz);        for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)for(int x=-B;x<=B;x++)if(f[j][i][x+B]<inf)            for(int k=0;k<=n;k++)for(int y=max(-B,-B-x);y<=B&&x+y<=B;y++)if(f[i][k][y+B]<inf)                up(f[j][k][x+y+B],f[j][i][x+B]+f[i][k][y+B]);        for(int i=0;i<=n;i++)for(int j=0;j<=2*B;j++){            if(j==B){if(f[i][i][j]<0){puts("0");goto be;}}            else if(f[i][i][j]<inf)j>B?l=max(l,(-f[i][i][j]-1)/(j-B)+1):r=min(r,-f[i][i][j]/(j-B));        }printf("%lld\n",r==inf?-1:(l<=r?r-l+1:0));be:;    }}

 

51nod 1340 差分约束