HDU 4738 Caocao's Bridges（雙聯通分量+並查集）

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大意：有n座島和m條橋，每條橋上有w個兵守著，現在要派不少於守橋計程車兵數的人去炸橋，只能炸一條橋，使得這n座島不連通，求最少要派多少人去。

思路：我們就是要縮點後直接求橋上人的最少數量。（PS：1、注意圖如果不聯通直接輸出0。2、如果圖中的橋上人為0，個那麼要讓一個人去。3、重邊的問題。這裡可以忽略）

#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（雙聯通分量+並查集）