HDU 4738 Caocao ' s Bridges (double Unicom component + and check set)

Effect: there are N islands and M-Bridge, each bridge has a W soldiers guarding, now to send not less than the number of soldiers to defend the bridge to the bridge, can only fry a bridge, so that the N island is not connected, ask at least how many people to send.

Idea: We just need to get the minimum number of the bridge directly after shrinking. (ps:1, note the figure if the direct output 0 is not unicom.) 2, if the figure of the bridge is 0, one so let a person go. 3, the problem of heavy edge. can be ignored here)

#include <map> #include <queue> #include <cmath> #include <cstdio> #include <stack># include<iostream> #include <cstring> #include <algorithm> #define LL int#define inf 0x3f3f3f3f#define EPS 1e-8#include<vector> #define LS l,mid,rt<<1#define rs mid+1,r,rt<<1|1using namespace Std;const int Ma = 1100;struct node{int to,w,next;} Q[ma*ma];int Head[ma*ma],dfn[ma],num[ma],du[ma],stk[ma*5],vis[ma],low[ma];int Cnt,top,tim,scc,out[Ma],f[Ma],n,mi    void Add (int a,int b,int c) {q[cnt].to = b;    Q[CNT].W = C;    Q[cnt].next = Head[a]; Head[a] = cnt++;}    void Init () {SCC = CNT = top = 0;    Tim = 1;    MI = inf;    memset (head,-1,sizeof (head));        for (int i = 1;i <= n;++ i) {f[i] = i;    Low[i] = vis[i] = Out[i] = Num[i] = Dfn[i] = 0;    }}void Tarjan (int u,int to) {Low[u] = dfn[u] = tim++;    Vis[u] = 1;    stk[top++] = u;        for (int i = head[u]; ~i; i = q[i].next) {int v = q[i].to; if (i = = (to^1)) ContinuE            if (!vis[v]) {Tarjan (v,i);            Low[u] = min (low[u],low[v]);        if (Low[v] > Dfn[u]) if (Q[I].W < mi) mi = q[i].w;    } else Low[u] = min (low[u],dfn[v]);        } if (low[u] = = Dfn[u]) {scc++;            while (Top > 0&&stk[top]! = u) {top--;            Vis[stk[top]] = 1;        Num[stk[top]] = SCC; }}}int Fi (int x) {return f[x] = = x x:f[x]=fi (f[x]);}    void mer (int a,int b) {int x = fi (a);    int y = fi (b); x > y? F[X] = y:f[y] = x;}    int main () {int m,i,j,k,a,b,c,cla;        while (~SCANF ("%d%d", &n,&m)) {if (!n&&!m) break;        Init ();                for (i = 0;i < m;++ i) {scanf ("%d%d%d", &a,&b,&c);                ADD (A,B,C);                ADD (B,A,C);        Mer (A, b);        } int tmp = 0;                for (i = 1;i <= n;++ i) if (f[i] = = i) {tmp++; if (tMP > 1) break;        } if (tmp > 1) {puts ("0"); continue;        } Tarjan (1,-1);                for (i = 1;i <= n;++ i) {for (j = head[i]; ~j; j=q[j].next) {int v = q[j].to;                if (Num[i]!=num[v]) {out[num[i]]++;        }}} int ans = 0;        for (i = 1;i <= scc;++ i) {if (out[i]==0) ans++;        } if (ans==1) {puts ("-1");        } else if (!mi) {puts ("1");    } else printf ("%d\n", MI); } return 0;}

HDU 4738 Caocao ' s Bridges (double Unicom component + and check set)