• HDU 3416 Marriage Match IV(ISAP+最短路)题解


    题意:从A走到B,有最短路,问这样不重复的最短路有几条

    思路:先来讲选有效边,我们从start和end各跑一次最短路,得到dis1和dis2数组,如果dis1[u] + dis2[v] + cost[u][v] == dis1[end],那么uv这条边是最短路的一条边。然后我们选完边,把边加入ISAP,然后跑一边就行了...还没学过SAP只会敲模板....

    错误思路:刚开始想的是先求出最短路,然后用费用流spfa去跑,边容量1,如果跑出一条路径费用等于最短路,那么路径+1,继续跑,但是超时了,看了半天是spfa跑费用流太慢的关系...等我学会网络流其他算法再来看这种emmm

    代码:

    #include<cstdio>
    #include<set>
    #include<cmath>
    #include<stack>
    #include<vector>
    #include<queue>
    #include<cstring>
    #include<string>
    #include<sstream>
    #include<iostream>
    #include<algorithm>
    #define ll long long
    using namespace std;
    const int maxn = 1000+5;
    const int INF = 0x3f3f3f3f;
    
    //网络流ISAP
    struct Node{
        int to,next,cap,flow;
    }edge[200005];
    int tot;
    int head[maxn];
    int gap[maxn],dep[maxn],pre[maxn],cur[maxn];
    void init(){
        tot = 0;
        memset(head,-1,sizeof(head));
    }
    void addEdge(int u,int v,int w,int rw = 0){
        edge[tot].to = v;
        edge[tot].cap = w;
        edge[tot].flow = 0;
        edge[tot].next = head[u];
        head[u] = tot++;
    
        edge[tot].to = u;
        edge[tot].cap = rw;
        edge[tot].flow = 0;
        edge[tot].next = head[v];
        head[v] = tot++;
    }
    int sap(int start,int end,int N){
        memset(gap,0,sizeof(gap));
        memset(dep,0,sizeof(dep));
        memcpy(cur,head,sizeof(head));
        int u = start;
        pre[u] = -1;
        gap[u] = N;
        int ans = 0;
        while(dep[start] < N){
            if(u == end){
                int Min = INF;
                for(int i = pre[u];i != -1;i = pre[edge[i^1].to]){
                    if(Min > edge[i].cap - edge[i].flow){
                        Min = edge[i].cap - edge[i].flow;
                    }
                }
                for(int i = pre[u];i != -1;i = pre[edge[i^1].to]){
                    edge[i].flow += Min;
                    edge[i^1].flow -= Min;
                }
                u = start;
                ans += Min;
                continue;
            }
            bool flag = false;
            int v;
            for(int i = cur[u];i != -1;i = edge[i].next){
                v = edge[i].to;
                if(edge[i].cap- edge[i].flow && dep[v] + 1 == dep[u]){
                    flag = true;
                    cur[u] = pre[v] = i;
                    break;
                }
            }
            if(flag){
                u = v;
                continue;
            }
            int Min = N;
            for(int i = head[u];i != -1;i = edge[i].next){
                if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){
                    Min = dep[edge[i].to];
                    cur[u] = i;
                }
            }
            gap[dep[u]]--;
            if(!gap[dep[u]]) return ans;
            dep[u] = Min + 1;
            gap[dep[u]]++;
            if(u != start) u = edge[pre[u]^1].to;
        }
        return ans;
    }
    
    struct road{
        int v,cost,next;
    }e[200005];
    int head2[maxn],tol,MinPath;
    void addRoad(int u,int v,int w){
        e[tol].v = v;
        e[tol].cost = w;
        e[tol].next = head2[u];
        head2[u] = tol++;
    }
    bool vis1[maxn];
    int cnt[maxn],dist[maxn];
    bool SPFA(int st,int n){
        memset(vis1,false,sizeof(vis1));
        for(int i = 0;i <= n;i++) dist[i] = INF;
        vis1[st] = true;
        dist[st] = 0;
        queue<int> q;
        while(!q.empty()) q.pop();
        q.push(st);
        memset(cnt,0,sizeof(cnt));
        cnt[st] = 1;
        while(!q.empty()){
            int u = q.front();
            q.pop();
            vis1[u] = false;
            for(int i = head2[u];i != -1;i = e[i].next){
                int v = e[i].v;
                if(dist[v] > dist[u] + e[i].cost){
                    dist[v] = dist[u] + e[i].cost;
                    if(!vis1[v]){
                        vis1[v] = true;
                        q.push(v);
                        if(++cnt[v] > n) return false;
                    }
                }
            }
        }
        return true;
    }
    int dis1[maxn];
    int u[100005],v[100005],w[100005];
    int main(){
        int n,m,T;
        scanf("%d",&T);
        while(T--){
            init();
            scanf("%d%d",&n,&m);
            for(int i = 0;i < m;i++){
                scanf("%d%d%d",&u[i],&v[i],&w[i]);
            }
            int st,en;
            scanf("%d%d",&st,&en);
            tol = 0;
            memset(head2,-1,sizeof(head2));
            for(int i = 0;i < m;i++){
                addRoad(u[i],v[i],w[i]);
            }
            SPFA(st,n);
            MinPath = dist[en];
            memcpy(dis1,dist,sizeof(dist));
            tol = 0;
            memset(head2,-1,sizeof(head2));
            for(int i = 0;i < m;i++){
                addRoad(v[i],u[i],w[i]);
            }
            SPFA(en,n);
            for(int i = 0;i < m;i++){
                if(dis1[u[i]] + dist[v[i]] + w[i] == MinPath)
                    addEdge(u[i],v[i],1);
            }
            int flow = sap(st,en,n);
            printf("%d
    ",flow);
        }
        return 0;
    }
  • 相关阅读:
    狗狗对主人的十句话
    微软提供正确卸载IE7的方法并恢复IE6
    c语言操作符的优先级
    linux 常用命令每日更新
    Visual Studio快捷键大全
    Opera将尽快发布补丁修复桌面浏览器漏洞 狼人:
    KILL杀毒软件重出江湖 公司股权全内资组成 狼人:
    图文:2010中国计算机网络安全年会华为展台 狼人:
    微软再次警告IE安全漏洞成为攻击目标 狼人:
    微软发布三月安全公告 两个补丁修补严重漏洞 狼人:
  • 原文地址:https://www.cnblogs.com/KirinSB/p/9479427.html
Copyright © 2020-2023  润新知