【HDU5889】Barricade（最短路+最小割）

记录一个菜逼的成长。。

题目大意：
给你N个点，M条边。让你在每条最短路上设置至少一个障碍使得路无法通过，每一个障碍需要Wi个木材，求需要最少的木材。

显然只要每条最短路设置一个障碍，只要求最短路的最小割即可。
直接套模板。。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <deque>
#include <cctype>
#include <bitset>
#include <cmath>
using namespace std;
#define ALL(v) (v).begin(),(v).end()
#define cl(a,b) memset(a,b,sizeof(a))
#define bp __builtin_popcount
#define pb push_back
#define fin freopen("D://in.txt","r",stdin)
#define fout freopen("D://out.txt","w",stdout)
#define lson t<<1,l,mid
#define rson t<<1|1,mid+1,r
#define seglen (node[t].r-node[t].l+1)
#define pi 3.1415926
#define exp  2.718281828459
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef vector<PII> VPII;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename T>
inline void read(T &x){
    T ans=0;
    char last=' ',ch=getchar();
    while(ch<'0' || ch>'9')last=ch,ch=getchar();
    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
    if(last=='-')ans=-ans;
    x = ans;
}
inline bool DBread(double &num)
{
    char in;double Dec=0.1;
    bool IsN=false,IsD=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))
        in=getchar();
    if(in=='-'){IsN=true;num=0;}
    else if(in=='.'){IsD=true;num=0;}
    else num=in-'0';
    if(!IsD){
        while(in=getchar(),in>='0'&&in<='9'){
            num*=10;num+=in-'0';}
    }
    if(in!='.'){
        if(IsN) num=-num;
            return true;
    }else{
        while(in=getchar(),in>='0'&&in<='9'){
                num+=Dec*(in-'0');Dec*=0.1;
        }
    }
    if(IsN) num=-num;
    return true;
}
template <typename T>
inline void write(T a) {
    if(a < 0) { putchar('-'); a = -a; }
    if(a >= 10) write(a / 10);
    putchar(a % 10 + '0');
}
/******************head***********************/
const int maxn = 1000 + 10;
const int MAX_V = maxn;
int g[maxn][maxn],val[maxn][maxn];
int dis[maxn],vis[maxn];
//dijkstra
void init(int n)
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            g[i][j]=INF;
        }
        dis[i]=INF;
    }
}
void dijkstra(int s,int n)
{
    dis[s]=0;
    for(int i=1;i<n;i++){
        int mn=INF,x;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&dis[j]<mn)mn=dis[x=j];
        }
        vis[x]=1;
        for(int j=1;j<=n;j++){
            dis[j]=min(dis[j],dis[x]+g[x][j]);
        }
    }
}
//Dinic求最大流，最大流等同最小割。
struct edge{
    int to,cap,rev;
    edge(){}
    edge(int _to,int _cap,int _rev):to(_to),cap(_cap),rev(_rev){}
};
vector<edge>G[MAX_V];
int level[MAX_V];
int iter[MAX_V];
void add(int from,int to,int cap)
{
    G[from].push_back(edge(to,cap,G[to].size()));
    G[to].push_back(edge(from,0,G[from].size()-1));
}
void bfs(int s)
{
    memset(level,-1,sizeof(level));
    queue<int>que;
    level[s] = 0;
    que.push(s);
    while(!que.empty()){
        int f = que.front();
        que.pop();
        for( int i = 0; i < G[f].size(); i++ ){
            edge &e = G[f][i];
            if(e.cap > 0 && level[e.to] == -1){
                level[e.to] = level[f] + 1;
                que.push(e.to);
            }
        }
    }
}
int dfs(int v,int t,int f)
{
    if(v == t)return f;
    for( int &i = iter[v]; i < G[v].size(); i++ ){
        edge &e = G[v][i];
        if(e.cap > 0 && level[v] < level[e.to]){
            int d = dfs(e.to,t,min(e.cap,f));
            if(d > 0){
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}
int max_flow(int s,int t)
{
    int flow = 0;
    for(;;){
        bfs(s);
        if(level[t] == -1)return flow;
        memset(iter,0,sizeof(iter));
        int f;
        while((f = dfs(s,t,INF)) > 0)
            flow += f;
    }
}
int main()
{
    int T;scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        init(n);
        for( int i = 1; i <= n; i++ )G[i].clear();
        for( int w,u,v,i = 0; i < m; i++ ){
            scanf("%d%d%d",&u,&v,&w);
            g[u][v] = 1;
            g[v][u] = 1;
            val[u][v] = w;
            val[v][u] = w;
        }
        dijkstra(1,n);
        for( int i = 1; i <= n; i++ ){
            for( int j = 1; j <= n; j++ ){
                //判断（i,j）这条边是否是在最短路中
                if(g[i][j] == 1 && dis[i] + 1 == dis[j]){
                    add(i,j,val[i][j]);
                }
            }
        }
        printf("%d\n",max_flow(1,n));
    }
    return 0;
}