bzoj 1927: [Sdoi2010]星际竞速

题目链接

bzoj 1927: [Sdoi2010]星际竞速

题解

可以考虑最小路径覆盖
转化成二分图之后进行最小费用最大流，把点拆违(x,x^{'})对于连向的边连到(x')上，对于连出点(u)只能经过u一次,流量为1,花费为cost[u->v]，(S->x,x'->T)
也可进行无源汇上下界网络流
拆点限制最小流量下线为1，建一条流量为1的边，比上一种方法多2n条边，但是，快一些(来自某dalao测试)


此处只给出第一种解法code

代码

/**************************************************************
    Problem: 1927
    User: 11101001
    Language: C++
    Result: Accepted
    Time:3572 ms
    Memory:2648 kb
****************************************************************/
 
// luogu-judger-enable-o2
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using std::max;
using std::min;
using std::queue;
const int maxm = 85007;
const int maxn = 8007;
int n,m,S=0,T;
inline int read() {
    int x=0;
    char c=getchar();
    while(c<'0'||c>'9') c=getchar();
    while(c<='9'&&c>='0') x=x*10+c-'0',c=getchar();
    return x;
}
struct Edge {
    int u,v,next,flow,cost;
}edge[maxm];
int a[maxn];
int num=1,head[maxn];
void add_edge(int u,int v,int flow,int cost) {
    edge[++num].u=u,edge[num].v=v,edge[num].flow=flow;edge[num].cost = cost;edge[num].next = head[u];head[u]=num;
    edge[++num].u=v;edge[num].v=u;edge[num].flow=0,edge[num].cost= -cost;edge[num].next = head[v];head[v]=num;
}
int dis[maxn],pre[maxn];bool vis[maxn];
bool spfa() {
    memset(dis,0x3f,sizeof dis);
    vis[S]=1;dis[S]=0;
    queue<int>que;
    que.push(S);
    while(!que.empty()) {
        int u=que.front();
        que.pop();
        for(int i=head[u];i;i=edge[i].next) {
            int v=edge[i].v;
            if(edge[i].flow&&dis[u]+edge[i].cost<dis[v]) {
                dis[v]=edge[i].cost+dis[u];
                pre[v]=i;
                if(!vis[v]) 
                    que.push(v),vis[v]=1;
            }
        }
        vis[u]=0;
    }
    if(dis[T]!=0x3f3f3f3f)return true;
    else return false;
}
int calc() {
    int ret=0;
    int maxx = 0x3f3f3f3f;
    for(int i=T;i!=S;i=edge[pre[i]].u)
        maxx=min(edge[pre[i]].flow,maxx);
    for(int i=T;i!=S;i=edge[pre[i]].u) {
        edge[pre[i]].flow-=maxx;
        edge[pre[i]^1].flow+=maxx;
        ret+=maxx*edge[pre[i]].cost;
    }
    return ret;
}
int ans=0;
void HowToMakeName() {
    while(spfa())
        ans+=calc();
}
int main() {
    n=read(),m=read();T=n<<1|1;
    for(int cost,i=1;i<=n;++i) {
        cost=read(),
        add_edge(S,n+i,1,cost);
        add_edge(S,i,1,0);
        add_edge(n+i,T,1,0);
    }
    for(int a,b,c,i=1;i<=m;++i) {
        a=read(),b=read(),c=read();
        if(a>b)std::swap(a,b);
        add_edge(a,b+n,1,c);
    }
    HowToMakeName();
    printf("%d
",ans);
    return 0;
}