zoj 2676 Network Wars(最小割，01分數規劃)

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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2676

大致題意：給出一個帶權無向圖，每條邊有一個邊權wi,求將S和T分開的一個割邊集C，使得該割邊集的平均邊權最小，即最小化∑wi / |C| 。

詳見amber關於最小割模型的論文

思路：amber論文中詳細講解了如何轉化成函數及建圖，值得注意的是當邊被重新賦權後,對於wi < 0 的邊權，該邊必然在最小割中，不必再建邊，直接加入最大流中即可，因為求最小割時邊權都為正值。

最後輸出的是所選割邊的序號。求割邊無非是從源點dfs,每次走殘量網路中流量大於0的邊並標記端點，最後判斷邊的兩個端點一個標記一個未標記，那麼該邊便是割邊。

這題我TLE了13次，最後是因為Dinic的原因，可能之前的那個模板耗時太長了。改成了學長的Dinic,瞬間就A了。

#include <stdio.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#define LL long long#define _LL __int64#define eps 1e-7using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 210;const int maxm = 6000;int s,t;struct node{    int u,v;    double w;    int next;    int re;}p[maxm],edge[maxm];int n,m;int cnt,head[maxn];int dist[maxn],vis[maxn];void init(){    cnt = 0;    memset(head,-1,sizeof(head));}void add(int u, int v, double w){    edge[cnt] = (struct node){u,v,w,head[u],cnt+1};    head[u] = cnt++;    edge[cnt] = (struct node){v,u,0,head[v],cnt-1};    head[v] = cnt++;}bool bfs(){    queue <int> que;    memset(dist, 0, sizeof(dist));    memset(vis, 0, sizeof(vis));    while(!que.empty()) que.pop();    vis[s] = 1;    que.push(s);    while(!que.empty())    {        int u = que.front();        que.pop();        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].v;            if(edge[i].w && !vis[v])            {                que.push(v);                vis[v] = 1;                dist[v] = dist[u]+1;            }        }    }    if(dist[t] == 0)return false;return true;}double dfs(int u, double delta){    if(u == t) return delta;    double ret = 0,tmp;for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(edge[i].w && dist[edge[i].v] == dist[u]+1 && (tmp = dfs(v,min(delta,edge[i].w)))){edge[i].w -= tmp;edge[edge[i].re].w += tmp;return tmp;}}if(!ret) dist[u] = -1;return ret;}double Dinic(){    double ret = 0,res;    while(bfs())    {       while(res = dfs(s,INF))ret += res;    }    return ret;}bool ok(double mid){    init();    double flow = 0;    for(int i = 1; i <= m; i++)    {        if(p[i].w > mid)        {            add(p[i].u,p[i].v,p[i].w-mid);            add(p[i].v,p[i].u,p[i].w-mid);        }        else            flow += p[i].w - mid;    }    flow += Dinic();    if(flow > eps)        return true;    else return false;}void dfs_cut(int u){    vis[u] = 1;    for(int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].v;        if(!vis[v] && edge[i].w > 0)        {            dfs_cut(v);        }    }}int main(){    int item = 0;    while(~scanf("%d %d",&n,&m))    {    s = 1;    t = n;        item += 1;        if(item >= 2)            printf("\n");        double low = INF,high = 0,mid;        for(int i = 1; i <= m; i++)        {            scanf("%d %d %lf",&p[i].u, &p[i].v, &p[i].w);            low = min(low,p[i].w);            high = max(high,p[i].w);        }        while( fabs(high - low) > eps)        {            mid = (high + low)/2.0;            if( ok(mid) )                low = mid;            else high = mid;        }        memset(vis,0,sizeof(vis));        dfs_cut(1);        int count = 0;        int ans[maxm];        for(int i = 1; i <= m; i++)        {            if(vis[p[i].u] + vis[p[i].v] == 1 || p[i].w <= mid)                ans[++count] = i;        }        printf("%d\n",count);        for(int i = 1; i <= count-1; i++)            printf("%d ",ans[i]);        printf("%d\n",ans[count]);    }    return 0;}