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  • BZOJ 1266 上学路线(最短路+最小割)

    给出n个点的无向图,每条边有两个属性,边权和代价。

    第一问求1-n的最短路。第二问求用最小的代价删边使得最短路的距离变大。

    对于第二问。显然该删除的是出现在最短路径上的边。如果我们将图用最短路跑一遍预处理出所有最短路径。

    然后我们要删除的边集一定是这个图的一个割。否则最短路径不会增加。即求此图的最小割。

    # include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi acos(-1.0)
    # define eps 1e-9
    # define MOD 1000000000
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=505;
    //Code begin...
    
    struct Edge{int p, next, w, d;}edge[N*N];
    struct Edge1{int p, next, w;}edge1[N*N];
    int head[N], head1[N], dist[N], cnt=1, cnt1=2, n, m, s, t;
    struct qnode{
        int v, c;
        qnode(int _v=0, int _c=0):v(_v),c(_c){}
        bool operator<(const qnode &r)const{return c>r.c;}
    };
    int vis[N];
    priority_queue<qnode>que;
    queue<int>Q;
    
    void add_edge(int u, int v, int d, int w){
        edge[cnt].p=v; edge[cnt].w=w; edge[cnt].d=d; edge[cnt].next=head[u]; head[u]=cnt++;
        edge[cnt].p=u; edge[cnt].w=w; edge[cnt].d=d; edge[cnt].next=head[v]; head[v]=cnt++;
    }
    void add_edge1(int u, int v, int w){
        edge1[cnt1].p=v; edge1[cnt1].w=w; edge1[cnt1].next=head1[u]; head1[u]=cnt1++;
        edge1[cnt1].p=u; edge1[cnt1].w=0; edge1[cnt1].next=head1[v]; head1[v]=cnt1++;
    }
    void Dijkstra(int n, int start){
        mem(vis,0); FOR(i,1,n) dist[i]=INF;
        dist[start]=0; que.push(qnode(start,0));
        qnode tmp;
        while (!que.empty()) {
            tmp=que.top(); que.pop();
            int u=tmp.v;
            if (vis[u]) continue;
            vis[u]=true;
            for (int i=head[u]; i; i=edge[i].next) {
                int v=edge[i].p, cost=edge[i].d;
                if (!vis[v]&&dist[v]>dist[u]+cost) dist[v]=dist[u]+cost, que.push(qnode(v,dist[v]));
            }
        }
    }
    int bfs(){
        int i, v;
        mem(vis,-1);
        vis[s]=0; Q.push(s);
        while (!Q.empty()) {
            v=Q.front(); Q.pop();
            for (i=head1[v]; i; i=edge1[i].next) {
                if (edge1[i].w>0 && vis[edge1[i].p]==-1) {
                    vis[edge1[i].p]=vis[v] + 1;
                    Q.push(edge1[i].p);
                }
            }
        }
        return vis[t]!=-1;
    }
    int dfs(int x, int low){
        int i, a, temp=low;
        if (x==t) return low;
        for (i=head1[x]; i; i=edge1[i].next) {
            if (edge1[i].w>0 && vis[edge1[i].p]==vis[x]+1){
                a=dfs(edge1[i].p,min(edge1[i].w,temp));
                temp-=a; edge1[i].w-=a; edge1[i^1].w+=a;
                if (temp==0) break;
            }
        }
        if (temp==low) vis[x]=-1;
        return low-temp;
    }
    int main ()
    {
        int u, v, d, w;
        scanf("%d%d",&n,&m);
        FOR(i,1,m) scanf("%d%d%d%d",&u,&v,&d,&w), add_edge(u,v,d,w);
        Dijkstra(n,1);
        printf("%d
    ",dist[n]);
        mem(vis,0); Q.push(n); vis[n]=true;
        while (!Q.empty()) {
            int tmp=Q.front(); Q.pop();
            for (int i=head[tmp]; i; i=edge[i].next) {
                int v=edge[i].p;
                if (dist[v]+edge[i].d!=dist[tmp]) continue;
                add_edge1(v,tmp,edge[i].w);
                if (!vis[v]) vis[v]=true, Q.push(v);
            }
        }
        s=1; t=n;
        int tmp, sum=0;
        while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp;
        printf("%d
    ",sum);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6759673.html
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