Codeforces Round #287 (Div. 2) E. Breaking Good 最短路

题目链接：

http://codeforces.com/problemset/problem/507/E

E. Breaking Good

time limit per test2 seconds
memory limit per test256 megabytes #### 问题描述 > Breaking Good is a new video game which a lot of gamers want to have. There is a certain level in the game that is really difficult even for experienced gamers. > > Walter William, the main character of the game, wants to join a gang called Los Hermanos (The Brothers). The gang controls the whole country which consists of n cities with m bidirectional roads connecting them. There is no road is connecting a city to itself and for any two cities there is at most one road between them. The country is connected, in the other words, it is possible to reach any city from any other city using the given roads. > > The roads aren't all working. There are some roads which need some more work to be performed to be completely functioning. > > The gang is going to rob a bank! The bank is located in city 1. As usual, the hardest part is to escape to their headquarters where the police can't get them. The gang's headquarters is in city n. To gain the gang's trust, Walter is in charge of this operation, so he came up with a smart plan. > > First of all the path which they are going to use on their way back from city 1 to their headquarters n must be as short as possible, since it is important to finish operation as fast as possible. > > Then, gang has to blow up all other roads in country that don't lay on this path, in order to prevent any police reinforcements. In case of non-working road, they don't have to blow up it as it is already malfunctional. > > If the chosen path has some roads that doesn't work they'll have to repair those roads before the operation. > > Walter discovered that there was a lot of paths that satisfied the condition of being shortest possible so he decided to choose among them a path that minimizes the total number of affected roads (both roads that have to be blown up and roads to be repaired). > > Can you help Walter complete his task and gain the gang's trust? #### 输入 > The first line of input contains two integers n, m (2 ≤ n ≤ 105, ), the number of cities and number of roads respectively. > > In following m lines there are descriptions of roads. Each description consists of three integers x, y, z (1 ≤ x, y ≤ n, ) meaning that there is a road connecting cities number x and y. If z = 1, this road is working, otherwise it is not. #### 输出 > In the first line output one integer k, the minimum possible number of roads affected by gang. > > In the following k lines output three integers describing roads that should be affected. Each line should contain three integers x, y, z (1 ≤ x, y ≤ n, ), cities connected by a road and the new state of a road. z = 1 indicates that the road between cities x and y should be repaired and z = 0 means that road should be blown up. > > You may output roads in any order. Each affected road should appear exactly once. You may output cities connected by a single road in any order. If you output a road, it's original state should be different from z. > > After performing all operations accroding to your plan, there should remain working only roads lying on some certain shortest past between city 1 and n. > > If there are multiple optimal answers output any. > ####样例输入 > 8 9 > 1 2 0 > 8 3 0 > 2 3 1 > 1 4 1 > 8 7 0 > 1 5 1 > 4 6 1 > 5 7 0 > 6 8 0

样例输出

3
2 3 0
1 5 0
6 8 1

题意

给你n个点，m条边的图，每条边长度都为1，如果标记为0，则这条边待修，1则可正常使用，现在让你找一条最短路，路径上面的待修理的边最少。

题解

跑完最短路之后建最短路构成的DAG图，然后跑拓扑排序跑dp。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;

struct Edge{
    int u,v,z;
    Edge(int u,int v,int z):u(u),v(v),z(z){}
};

struct Spfa{
    int n,m;
    vector<Edge> egs;
    VI G[maxn],G2[maxn];
    bool inq[maxn];
    int d[maxn],d2[maxn];

    void init(int n){
        this->n=n;
        for(int i=0;i<n;i++) G[i].clear();
        egs.clear();
    }

    void addEdge(int u,int v,int z){
        egs.pb(Edge(u,v,z));
        m=egs.sz();
        G[u].pb(m-1);
    }

    int spfa(int s,int *d){
        queue<int> Q;
        clr(inq,0);
        rep(i,0,n) d[i]=INF;
        d[s]=0,inq[s]=true,Q.push(s);
        while(!Q.empty()){
            int u=Q.front(); Q.pop();
            inq[u]=false;
            for(int i=0;i<G[u].sz();i++){
                Edge& e=egs[G[u][i]];
                if(d[e.v]>d[u]+1){
                    d[e.v]=d[u]+1;
                    if(!inq[e.v]){
                        Q.push(e.v);
                        inq[e.v]=true;
                    }
                }
            }
        }
    }

    int ind[maxn],dp[maxn];
    int pre[maxn];
    bool vis[maxn*2];
    void solve(){
        clr(pre,-1);
        clr(ind,0);
        clr(vis,0);
        clr(dp,0x3f);

        spfa(0,d);
        spfa(n-1,d2);

        for(int i=0;i<m;i++){
            Edge &e=egs[i];
            if(d[e.u]+1+d2[e.v]==d[n-1]){
//                    prf("(%d,%d)\n",e.u+1,e.v+1);
                G2[e.u].pb(i);
                ind[e.v]++;
            }
        }

        //
        queue<int> Q;
        Q.push(0),dp[0]=0;
        while(!Q.empty()){
            int u=Q.front(); Q.pop();
            rep(i,0,G2[u].sz()){
                Edge& e=egs[G2[u][i]];
                if(dp[e.v]>dp[u]+(e.z^1)){
                    dp[e.v]=dp[u]+(e.z^1);
                    pre[e.v]=G2[u][i];
                }
                ind[e.v]--;
                if(ind[e.v]==0){
                    Q.push(e.v);
                }
            }
        }

//        bug(dp[n-1]);

        VI ans;
        int p=n-1;
        while(p){
            Edge& e=egs[pre[p]];
//            prf("<%d,%d>\n",e.u,e.v);
            if(e.z==0){
                ans.pb(pre[p]);
            }
            vis[pre[p]]=vis[pre[p]^1]=1;
            p=e.u;
        }

        for(int i=0;i<m;i++){
            if(vis[i]) continue;
            Edge& e=egs[i];
            if(e.z==1){
                ans.pb(i);
            }
            vis[i]=vis[i^1]=1;
        }

        prf("%d\n",ans.sz());
        rep(i,0,ans.sz()){
            Edge& e=egs[ans[i]];
            prf("%d %d %d\n",e.u+1,e.v+1,e.z^1);
        }
    }

}spfa;

int main() {
    int n,m;
    scf("%d%d",&n,&m);
    spfa.init(n);
    for(int i=0;i<m;i++){
        int u,v,z;
        scf("%d%d%d",&u,&v,&z); u--,v--;
        spfa.addEdge(u,v,z);
        spfa.addEdge(v,u,z);
    }
    spfa.solve();
    return 0;
}

//end-----------------------------------------------------------------------

其实。。spfa跑最短路的时候就能处理出0最少的最短路径了。。orz..上面的做法好蠢。。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;

struct Edge{
    int u,v,z;
    Edge(int u,int v,int z):u(u),v(v),z(z){}
};

struct Spfa{
    int n,m;
    vector<Edge> egs;
    VI G[maxn];
    bool inq[maxn];
    int d[maxn],dp[maxn];
    int pre[maxn];

    void init(int n){
        this->n=n;
        for(int i=0;i<n;i++) G[i].clear();
        egs.clear();
    }

    void addEdge(int u,int v,int z){
        egs.pb(Edge(u,v,z));
        m=egs.sz();
        G[u].pb(m-1);
    }

    int spfa(int s){
        clr(pre,-1);
        clr(dp,0x3f);
        queue<int> Q;
        clr(inq,0);
        rep(i,0,n) d[i]=INF;
        dp[s]=0;
        d[s]=0,inq[s]=true,Q.push(s);
        while(!Q.empty()){
            int u=Q.front(); Q.pop();
            inq[u]=false;
            for(int i=0;i<G[u].sz();i++){
                Edge& e=egs[G[u][i]];
                if(d[e.v]>d[u]+1||d[e.v]==d[u]+1&&dp[e.v]>dp[u]+(e.z^1)){
                    d[e.v]=d[u]+1;
                    dp[e.v]=dp[u]+(e.z^1);
                    pre[e.v]=G[u][i];
                    if(!inq[e.v]){
                        Q.push(e.v);
                        inq[e.v]=true;
                    }
                }
            }
        }
    }

    bool vis[maxn*2];
    void solve(){
        clr(vis,0);

        spfa(0);

        VI ans;
        int p=n-1;
        while(p){
            Edge& e=egs[pre[p]];
            if(e.z==0){
                ans.pb(pre[p]);
            }
            vis[pre[p]]=vis[pre[p]^1]=1;
            p=e.u;
        }

        for(int i=0;i<m;i++){
            if(vis[i]) continue;
            Edge& e=egs[i];
            if(e.z==1){
                ans.pb(i);
            }
            vis[i]=vis[i^1]=1;
        }

        prf("%d\n",ans.sz());
        rep(i,0,ans.sz()){
            Edge& e=egs[ans[i]];
            prf("%d %d %d\n",e.u+1,e.v+1,e.z^1);
        }
    }

}spfa;

int main() {
    int n,m;
    scf("%d%d",&n,&m);
    spfa.init(n);
    for(int i=0;i<m;i++){
        int u,v,z;
        scf("%d%d%d",&u,&v,&z); u--,v--;
        spfa.addEdge(u,v,z);
        spfa.addEdge(v,u,z);
    }
    spfa.solve();
    return 0;
}

//end-----------------------------------------------------------------------

代码