标签：

Tricks Device

8 9
1 2 2
2 3 2
2 4 1
3 5 3
4 5 4
5 8 1
1 6 2
6 7 5
7 8 1

2 6

题意：

n个点，m条边，构建有权无向图。

求出删去最少条边数可以使得图没有最短路径，以及删出最多条边使得图仍有最多条路径。

思路：

（1）最短路处理出最短路径图，若dis[v] = dist[u] + w(u,v)，则该路在最短路径中。把最短路的边加进最短路径图，边权值为1，在新图求最小割。转化成对立问题，求1 -> N的最大流。

（2）在跑最短路中用side数组记录最短路的最少边数，m - side[ n ]即为解。

最短路处理出最短路径图是本题的难点， 想了好久都没想明白，最后宇哥哥给讲了一下， 还是对spfa的理解不够深入。

最短路处理出最短路径图代码：

//重见图
void getmap(){
    for(int i = 1; i <= n; ++i)//枚举起点
    for(int j = head1[i]; j != -1 ; j = str[j].next){
        NODE E = str[j];
        if(dist[E.v] == dist[i] + E.w)//加上最短路上的边
            addedge(i, E.v, 1);
    }
}

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define maxn 2000+100
#define maxm 200000+100
#define INF 0x3f3f3f3f
using namespace std;
int m, n;
int head[maxn], cur[maxn], cnt;
int head1[maxn], cnt1;
int dist[maxn], vis[maxn];
int side[maxn];
struct node{
    int u, v, cap, flow, next;
};
node edge[maxm];

struct NODE{
    int u, v, w, next;
};

NODE str[maxm];

void initedge(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void initstr(){
    cnt1 = 0;
    memset(head1, -1, sizeof(head1));
}

void addedge(int u, int v, int w){
    node E1 = {u, v, w, 0, head[u]};
    edge[cnt] = E1;
    head[u] = cnt++;
    node E2 = {v, u, 0, 0, head[v]};
    edge[cnt] = E2;
    head[v] = cnt++;
}

void addstr(int u, int v, int w){
    NODE E1 = {u, v, w, head1[u]};
    str[cnt1] = E1;
    head1[u] = cnt1++;
    NODE E2 = {v, u, w, head1[v]};
    str[cnt1] = E2;
    head1[v] = cnt1++;
}

void spfa(){//求出最短路径 + 到达N最少经过的边数,边数存在 side里面
    queue<int>q;
    for(int i = 1; i <= n; ++i){
        vis[i] = 0;
        dist[i] = INF;
        side[i] = 0;
    }
    vis[1] = 1;
    dist[1] = 0;
    q.push(1);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head1[u]; i != -1; i = str[i].next){
            int v = str[i].v;
            int w = str[i].w;
            if(dist[v] > dist[u] + w){
                dist[v] = dist[u] + w;
                side[v] = side[u] + 1;
                if(!vis[v]){
                    vis[v] = 1;
                    q.push(v);
                }
            }
            else if(dist[v] == dist[u] + w)
                side[v] = min(side[v], side[u] + 1);
        }
    }
}
//重见图
void getmap(){
    for(int i = 1; i <= n; ++i)//枚举起点
    for(int j = head1[i]; j != -1 ; j = str[j].next){
        NODE E = str[j];
        if(dist[E.v] == dist[i] + E.w)//加上最短路上的边
            addedge(i, E.v, 1);
    }
}

bool BFS(int st, int ed){
    queue<int>q;
    memset(vis, 0, sizeof(vis));
    memset(dist, -1, sizeof(dist));
    vis[st] = 1;
    dist[st] = 0;
    q.push(st);
    while(!q.empty()){
        int u =q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next){
            node E = edge[i];
            if(!vis[E.v] && E.cap > E.flow){
                vis[E.v] = 1;
                dist[E.v] = dist[u] + 1;
                if(E.v == ed) return true;
                q.push(E.v);
            }
        }
    }
    return false;
}

int DFS(int x, int ed, int a){
    if(x == ed || a == 0)
        return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1; i = edge[i].next){
        node &E = edge[i];
        if(dist[E.v] == dist[x] + 1 && (f = DFS(E.v, ed, min(a, E.cap - E.flow))) > 0){
            E.flow += f;
            edge[i ^ 1].flow -= f;
            a -= f;
            flow += f;
            if(a == 0) break;
        }
    }
    return flow;
}

int maxflow(int st, int ed){
    int flowsum = 0;
    while(BFS(st, ed)){
        memcpy(cur, head, sizeof(head));
        flowsum += DFS(st, ed, INF);
    }
    return flowsum;
}

int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        initstr();
        for(int i = 1; i <= m; ++i){
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            addstr(u, v, w);
        }
        spfa();
        initedge();
        getmap();
        printf("%d %d\n", maxflow(1, n), m - side[n]);
    }
    return 0;
}

HDU 5294--Tricks Device【最小割 && 最短路处理，新建图】

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原文地址：http://blog.csdn.net/hpuhjh/article/details/47345361