sgu206

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首先很显然，生成树上的边权应减小，非生成树上的边权应增大。

令 xi=ci−di , yj=dj−cj (xi>=0,yj>=0)
若树边 i 在非树边 j 和生成树形成的环上，
有 di<=dj⟹ci−xi<=yj+cj⟹xi+yj>=ci−cj

建立二分图，其中 wi,j=ci−cj，KM 算法求最大权匹配，满足
Ai+Bj>=wi,j，即可满足该不等式。

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与 差分约束系统 有着异曲同工之妙 ^_^。

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#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>const int MAXN = 70, MAXM = 405, INF = 0x3f3f3f3f;int n, m; int a[MAXM], b[MAXM], c[MAXM]; int g[MAXN][MAXN];int nx, ny, w[MAXM][MAXM];int link[MAXM], slack[MAXM];int lx[MAXM], ly[MAXM];bool visx[MAXM], visy[MAXM];bool DFS(int fr,int a,int to,int id){    if(a == to) return true;    for(int i = 1; i <= n; i++)    {        if(i == fr || !g[a][i]) continue;        if(DFS(a, i, to, id))        {            int &newrc = w[g[a][i]][id];            newrc = std::max(newrc, c[g[a][i]] - c[id]);            return true;        }    }    return false;}void PreWork(){    scanf("%d%d",&n,&m);    for(int i = 1; i <= m; i++)    {         scanf("%d%d%d",&a[i],&b[i],&c[i]);         if(i < n) g[a[i]][b[i]] = g[b[i]][a[i]] = i;    }    for(int i = n; i <= m; i++)        DFS(0,a[i],b[i],i);    nx = m, ny = m;}bool find(int x){    visx[x] = true;    for(int y = 1; y <= ny; y++)        if(!visy[y])        {            int cal = lx[x] + ly[y] - w[x][y];            if(cal == 0)            {                visy[y] = true;                if(!link[y] || find(link[y]))                {                    link[y] = x;                    return true;                }            }            slack[y] = std::min(cal, slack[y]);         }    return false;   }void Solve(){    memset(lx,0,sizeof(lx));    memset(ly,0,sizeof(ly));    for(int i = 1; i <= nx; i++)        for(int j = 1; j <= ny; j++)            lx[i] = std::max(lx[i], w[i][j]);    for(int i = 1; i <= nx; i++)    {        memset(slack,INF,sizeof(slack));         while(true)        {            memset(visx,false,sizeof(visx));            memset(visy,false,sizeof(visy));            if(find(i)) break;            int d = INF;            for(int j = 1; j <= ny; j++)                if(!visy[j]) d = std::min(d, slack[j]);            for(int j = 1; j <= nx; j++)                if(visx[j])lx[j] -= d;            for(int j = 1; j <= ny; j++)                if(visy[j])ly[j] += d;                else  slack[j] -= d;                }    }}void Output(){    for(int i = 1; i < n; i++)        printf("%d\n",c[i]-lx[i]);    for(int i = n; i <= m; i++)        printf("%d\n",c[i]+ly[i]);}int main(){#ifndef ONLINE_JUDGE    freopen("sgu206.in","r",stdin);    freopen("sgu206.out","w",stdout);#endif    PreWork();    Solve();    Output();#ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout); #endif    return 0;   }