hdu4005The war

鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4005

題意：給定n個點m條邊的無向圖，求任意添加一條邊後的最小割邊的最大值。

分析：因爲是圖很明顯我們要先縮點將圖縮成一顆樹，然後會發現題目變成在樹上添加一條邊求最小不在環上的最大值。很顯然那個最大值應該儘量不和最小邊在一條鏈上，那麼我們以最小邊爲根dfs確定一條鏈儘可能將小邊包含使得不在鏈上的邊的最小值最大。

代碼：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=10010;
const int M=100010;
const int mod=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1004535809;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-5;
const double inf=~0u>>1;
const double pi=acos(-1.0);
typedef long double db;
typedef unsigned int uint;
typedef unsigned long long ull;
struct Edge {
    int u,v,w,id,nex;
    Edge() {}
    Edge(int u,int v,int w,int id,int nex):u(u),v(v),w(w),id(id),nex(nex) {}
}edge[M<<1];
int ans,tot,first[N];
int sccn,dfs_clock,pre[N],dfn[N],vis[M],sccno[N];
void dfs(int u) {
    pre[u]=dfn[u]=++dfs_clock;
    for (int i=first[u];~i;i=edge[i].nex)
    if (!vis[edge[i].id]) {
        vis[edge[i].id]=1;
        if (!pre[edge[i].v]) {
            dfs(edge[i].v);
            dfn[u]=min(dfn[u],dfn[edge[i].v]);
            if (dfn[edge[i].v]>pre[u]) vis[edge[i].id]=2;
        } else dfn[u]=min(dfn[u],pre[edge[i].v]);
    }
}
void dfs_num(int u) {
    sccno[u]=sccn;
    for (int i=first[u];~i;i=edge[i].nex)
    if (vis[edge[i].id]==1&&!sccno[edge[i].v]) {
        vis[edge[i].id]=0;dfs_num(edge[i].v);
    }
}
void dfs_scc(int n,int m) {
    sccn=dfs_clock=0;
    for (int i=1;i<=m;i++) vis[i]=0;
    for (int i=1;i<=n;i++) pre[i]=sccno[i]=0;
    for (int i=1;i<=n;i++)
    if (!pre[i]) dfs(i);
    for (int i=1;i<=n;i++)
    if (!sccno[i]) sccn++,dfs_num(i);
}
struct node {
    int v,w;
    node() {}
    node(int v,int w):v(v),w(w) {}
};
vector<node>V[N];
void dfs(int u,int v,int w) {
    int i;dfn[u]=0;
    for (i=0;i<V[u].size();i++)
    if (V[u][i].v!=v) {
        dfs(V[u][i].v,u,V[u][i].w);
        if (pre[V[u][i].v]<pre[dfn[u]]) dfn[u]=V[u][i].v;
    }
    pre[u]=min(w,pre[dfn[u]]);
}
void dfs_ans(int u,int v,int w,int bo) {
    if (!bo) ans=min(ans,w);
    for (int i=0;i<V[u].size();i++)
    if (V[u][i].v!=v) {
        if (V[u][i].v==dfn[u]) dfs_ans(V[u][i].v,u,V[u][i].w,bo);
        else dfs_ans(V[u][i].v,u,V[u][i].w,0);
    }
}
int main()
{
    int i,n,m,u,v,w,mi;
    while (scanf("%d%d", &n, &m)!=EOF) {
        for (tot=0,i=1;i<=n;i++) first[i]=-1;
        for (i=1;i<=m;i++) {
            scanf("%d%d%d", &u, &v, &w);
            edge[tot]=Edge(u,v,w,i,first[u]);first[u]=tot++;
            edge[tot]=Edge(v,u,w,i,first[v]);first[v]=tot++;
        }
        dfs_scc(n,m);
        u=v=0;mi=INF;
        for (i=1;i<=sccn;i++) V[i].clear();
        for (i=0;i<tot;i+=2)
        if (sccno[edge[i].u]!=sccno[edge[i].v]) {
            V[sccno[edge[i].u]].push_back(node(sccno[edge[i].v],edge[i].w));
            V[sccno[edge[i].v]].push_back(node(sccno[edge[i].u],edge[i].w));
            if (edge[i].w<mi) mi=edge[i].w,u=sccno[edge[i].u],v=sccno[edge[i].v];
        }
        ans=pre[0]=INF;
        dfs(u,v,INF);dfs(v,u,INF);
        dfs_ans(u,v,w,1);dfs_ans(v,u,w,1);
        if (ans==INF) printf("-1\n");
        else printf("%d\n", ans);
    }
    return 0;
}