CSES - Datatähti Open 2017 - Results
Submission details
Task:Tunnels
Sender:ainta1
Submission time:2017-01-22 05:59:44 +0200
Language:C++
Status:READY
Result:0
Feedback
groupverdictscore
#10
#20
#30
Test results
testverdicttimegroup
#1ACCEPTED0.04 s1details
#20.04 s1details
#30.05 s1details
#40.03 s1details
#50.05 s1details
#60.05 s2details
#70.04 s2details
#80.04 s2details
#90.04 s2details
#100.03 s2details
#110.09 s3details
#120.08 s3details
#130.05 s3details
#140.07 s3details
#150.04 s3details

Compiler report

input/code.cpp: In function 'int main()':
input/code.cpp:395:24: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf("%d%d",&n,&m);
                        ^
input/code.cpp:398:28: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
         scanf("%d%d",&a,&b);
                            ^

Code

/*#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
int n, m, RR[40][40], DD[40][40], ES[40][40];
long long res;
int Num(int x, int y, int ck){
    return (x-1)*m+y + ck*n*m;
}
struct Edge{
    int b, e, c, f;
}E[201000];
int EC, source, sink;
vector<int>G[5000];

void Make_Edge(int b, int e, int f, int c){
    G[b].push_back(EC);G[e].push_back(EC+1);
    E[EC].b=b,E[EC].e=e,E[EC].f=f,E[EC].c=c;EC++;
    E[EC].b=e,E[EC].e=b,E[EC].f=0,E[EC].c=-c;EC++;
}

int Path[10100], D[10100], Q[1010000], InQ[10100];

int SPFA(){
    int i, head = 0, tail = 0, x;
    for(i=0;i<=sink;i++){
        D[i] = -1e9;
        InQ[i] = 0;
    }
    D[0] = 0;
    Q[++tail] = 0;
    InQ[0] = 1;
    while(head < tail){
        x = Q[++head];
        InQ[x] = 0;
        for(i=0;i<G[x].size();i++){
            Edge tp = E[G[x][i]];
            if(tp.f && D[tp.e] < D[x] + tp.c){
                D[tp.e] = D[x] + tp.c;
                Path[tp.e] = G[x][i];
                if(!InQ[tp.e]){
                    InQ[tp.e] = 1;
                    Q[++tail] = tp.e;
                }
            }
        }
    }
    return D[sink];
}

void MCMF(){
    int t, x, y, f;
    while((t = SPFA()) >= 0){
        x = sink;
        f = 100;
        while(x){
            y = Path[x];
            f=min(f,E[y].f);
            x = E[y].b;
        }
        res += 1ll*f*t;
        x = sink;
        while(x){
            y = Path[x];
            E[y].f -= f;
            E[y^1].f += f;
            x = E[y].b;
        }
    }
}

int main(){
    int TC,TT,i,j,K;
    scanf("%d",&TC);
    for(TT=1;TT<=TC;TT++){
        printf("Case #%d: ",TT);
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)for(j=1;j<m;j++)scanf("%d",&RR[i][j]);
        for(i=1;i<n;i++)for(j=1;j<=m;j++)scanf("%d",&DD[i][j]);
        EC = 0;
        source = 0, sink = n*m*2+1;
        scanf("%d",&K);
        for(i=1;i<=K;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            ES[x][y]=1;
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
            }
        }
        
        MCMF();
        if(res < -5000000){
            printf("Impossible\n");
        }
        else{
            printf("%lld\n",res);
        }
        for(i=0;i<=sink;i++){
            G[i].clear();
        }
    }
}*/

/*
#include<stdio.h>
#include<algorithm>
#include<cstdlib>
using namespace std;
int n, K, w[1010000];
long long res, S[1010000], Sum, D1[1010000][8], D2[1010000][8], D[64][2], DP[2048][2];
bool v[64][2];
int C[2048], M[2048], m, Path[2048][2];
bool vv[2048][2];
void Dijk(){
    int i, j;
    for(i=0;i<(1<<m);i++)v[i][0]=v[i][1]=false;
    while(1){
        int x=-1, y=-1;
        long long MM = 1e17;
        for(i=0;i<(1<<m);i++){
            for(j=0;j<2;j++){
                if(MM > D[i][j] && !v[i][j]){
                    MM = D[i][j];
                    x=i,y=j;
                }
            }
        }
        if(x==-1)break;
        v[x][y]=true;
        if(y==0){
            for(j=0;j<(1<<m);j++){
                if(j&x || C[j]>3)continue;
                D[x^j][1] = min(D[x^j][1], D[x][y] + M[j]);
            }
        }
        else{
            for(j=0;j<m;j++){
                if(!((1<<j)&x))continue;
                D[x^(1<<j)][0] = min(D[x^(1<<j)][0], D[x][y] + w[j+1]);
            }
        }
    }
}
void Dijk2(){
    int i, j;
    for(i=0;i<(1<<n);i++)vv[i][0]=vv[i][1]=false;
    while(1){
        int x=-1, y=-1;
        long long MM = 1e17;
        for(i=0;i<(1<<n);i++){
            for(j=0;j<2;j++){
                if(MM > DP[i][j] && !vv[i][j]){
                    MM = DP[i][j];
                    x=i,y=j;
                }
            }
        }
        if(x==-1)break;
        vv[x][y]=true;
        if(y==0){
            for(j=0;j<(1<<n);j++){
                if(j&x || C[j]>3)continue;
                if(DP[x^j][1] > DP[x][y] + M[j]) Path[x^j][1] = x;
                DP[x^j][1] = min(DP[x^j][1], DP[x][y] + M[j]);
            }
        }
        else{
            for(j=0;j<n;j++){
                if(!((1<<j)&x))continue;
                if(DP[x^(1<<j)][0]> DP[x][y] + w[j+1]) Path[x^(1<<j)][0] = x;
                DP[x^(1<<j)][0] = min(DP[x^(1<<j)][0], DP[x][y] + w[j+1]);
            }
        }
    }
}
int TP[1010];
long long Do(int a, int b, int c){
    int t = a*3-a-b-c, i;
    long long S = 1ll*a*w[1]+1ll*b*w[2]+1ll*c*w[3];
    for(i=n;i>=6;i-=3){
        if(i-3 >= 3+t){
            S += w[i];
        }
        else break;
    }
    int ed = i;
    for(i=ed;i>=4;i--)TP[i] = 1;
    TP[1]=a+1,TP[2]=b+1,TP[3]=c+1;
    int s = ed+a+b+c;
    if(s%3==2){
        TP[1]--,TP[2]--;
        S+=w[2];
    }
    while(1){
        int x1=-1,x2=-1,x3=-1;
        for(i=1;i<=n;i++){
            if(TP[i]){
                if(x1==-1)x1=i;
                else if(x2==-1)x2=i;
                else if(x3==-1){
                    x3=i;
                    break;
                }
            }
        }
        if(x1==-1)break;
        TP[x1]--,TP[x2]--,TP[x3]--;
        S += w[x3];
    }
    return S;
}
int main(){
    int i, j, k;
    srand(1879);
    int tc=0;
    while(1){
        printf("%d\n",++tc);
        n = 9, K = 3;
        for(i=1;i<=n;i++){
            w[i] = rand()%5+1;
        }
        sort(w+1,w+n+1);
        for(i=0;i<=n;i++){
            for(j=0;j<8;j++){
                D1[i][j]=D2[i][j]=1e18;
            }
        }
        for(i=0;i<(1<<n);i++){
            C[i] = 0, M[i] = 0;
        }
        m = min(n,6);
        for(i=0;i<(1<<n);i++){
            for(j=0;j<n;j++)if(i&(1<<j))C[i]++, M[i] = w[j+1];
        }
        D1[n][0] = 0;
        res = 1e18;
        for(i=n;i>=6;i--){
            for(int cc = 0; cc <= 4; cc++){
                for(j=0;j<8;j++){
                    if(C[j] != cc)continue;
                    if(D2[i][j]>1e17)continue;
                    for(k=0;k<3;k++){
                        if((1<<k)&j){
                            D1[i][j^(1<<k)] = min(D1[i][j^(1<<k)], D2[i][j] + w[k+1]);
                        }
                    }
                }
                for(j=0;j<8;j++){
                    if(C[j] != cc-1)continue;
                    for(k=0;k<8;k++){
                        if(k&j)continue;
                        if(C[k]>=2)D2[i][j^k] = min(D2[i][j^k], D1[i][j] + M[k]);
                        if(C[k]==3)continue;
                        D2[i-1][j^k] = min(D2[i-1][j^k], D1[i][j] + w[i]);
                        if(C[k]==2)continue;
                        D2[i-2][j^k] = min(D2[i-2][j^k], D1[i][j] + w[i]);
                        if(C[k]==1)continue;
                        D2[i-3][j^k] = min(D2[i-3][j^k], D1[i][j] + w[i]);
                    }
                }
            }
        }
        for(i=0;i<(1<<m);i++)D[i][0]=D[i][1]=1e18;
        for(i=3;i<=m;i++){
            for(j=0;j<8;j++){
                int s = 0;
                for(k=0;k<m;k++){
                    if(k>=i || (k<3 && j&(1<<k))) s += (1<<k);
                }
                D[s][0] = D1[i][j];
                D[s][1] = D2[i][j];
            }
        }
        Dijk();
        long long res1 = D[(1<<m)-1][1];
        
        
        int ret = (n-2)/2;
        res1 = 1e18;
        for(i=0;i<=ret;i++){
            for(j=0;i+j<=ret;j++){
                k=ret-i-j;
                if(i<j||j<k)continue;
                res1 = min(res1,Do(i,j,k));
            }
        }
        
        for(i=0;i<(1<<n);i++)DP[i][0]=DP[i][1]=1e18;
        DP[0][0] = 0;
        Dijk2();
        long long res2 = DP[(1<<n)-1][1];
        if(res1 != res2){
            for(i=1;i<=n;i++)printf("%d ",w[i]);
            printf("\n");
            printf("%lld %lld\n",res1,res2);
            int x = (1<<n)-1, y = 1;
            while(x||y){
                printf("%d %d\n",x,y);
                x = Path[x][y];
                y = !y;
            }
            break;
        }
    }
}*/
/*
#include<stdio.h>
#include<algorithm>
using namespace std;
int n, K;
int w[1010000], TP[1010000];
long long S[1010000], res;
void Do(int a, int b, int c){
    int t = a*3-a-b-c, i;
    long long S = 1ll*a*w[1]+1ll*b*w[2]+1ll*c*w[3];
    for(i=n;i>=6;i-=3){
        if(i-3 >= 3+t){
            S += w[i];
        }
        else break;
    }
    int ed = i;
    for(i=ed;i>=4;i--)TP[i] = 1;
    TP[1]=a+1,TP[2]=b+1,TP[3]=c+1;
    int s = ed+a+b+c;
    if(s%3==2){
        TP[1]--,TP[2]--;
        S+=w[2];
    }
    while(1){
        int x1=-1,x2=-1,x3=-1;
        for(i=1;i<=n;i++){
            if(TP[i]){
                if(x1==-1)x1=i;
                else if(x2==-1)x2=i;
                else if(x3==-1){
                    x3=i;
                    break;
                }
            }
        }
        if(x1==-1)break;
        TP[x1]--,TP[x2]--,TP[x3]--;
        S += w[x3];
    }
    res = min(res,S);
}
int main(){
    int i, j, k;
    scanf("%d%d",&n,&K);
    for(i=1;i<=n;i++){
        scanf("%d",&w[i]);
        S[i] = S[i-1] + w[i];
    }
    if(n<=K){
        printf("%d\n",w[n]);
        return 0;
    }
    if(K==2){
        long long Sum;
        Sum = S[n]-S[1] + 1ll*w[1]*(n-2);
        res = Sum;
        for(i=n-1;i>=3;i-=2){
            Sum -= w[i]+w[i+1]+w[1]*2;
            Sum += w[i+1]+w[2]+w[2]+w[1];
            res = min(res,Sum);
        }
        printf("%lld\n",res);
        return 0;
    }
    if(n>100)return 0;
    int ret = (n-2)/2;
    res = 1e18;
    for(i=0;i<=ret;i++){
        for(j=0;i+j<=ret;j++){
            k=ret-i-j;
            if(i<j||j<k)continue;
            Do(i,j,k);
        }
    }
    printf("%lld\n",res);
}*/
#include<stdio.h>
#include<algorithm>
int UF[101000], n, m, Deg[101000], C[101000], chk[101000], res;
using namespace std;
int Find(int a){
    if(a==UF[a])return a;
    return UF[a] = Find(UF[a]);
}
int main(){
    int i, a, b;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)UF[i] = i;
    for(i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        Deg[a]++,Deg[b]--;
        a = Find(a), b = Find(b);
        if(a!=b){
            UF[a] = b;
        }
    }
    for(i=1;i<=n;i++){
        C[Find(i)] += (Deg[i]<0?-Deg[i]:Deg[i]);
        chk[Find(i)] = 1;
    }
    for(i=1;i<=n;i++){
        if(chk[i]){
            res += max(C[i],2)/2;
        }
    }
    printf("%d\n",res);
}

Test details

Test 1

Group: 1

Verdict: ACCEPTED

input
10 20
4 5
6 4
5 1
5 9
...

correct output
11

user output
11

Test 2

Group: 1

Verdict:

input
10 10
7 3
5 2
9 7
1 5
...

correct output
5

user output
6

Test 3

Group: 1

Verdict:

input
10 5
5 7
3 8
5 8
3 7
...

correct output
4

user output
9

Test 4

Group: 1

Verdict:

input
10 4
9 1
6 8
7 1
5 7

correct output
3

user output
7

Test 5

Group: 1

Verdict:

input
10 2
10 6
2 1

correct output
2

user output
8

Test 6

Group: 2

Verdict:

input
100 200
24 40
25 6
36 93
92 90
...

correct output
97

user output
102

Test 7

Group: 2

Verdict:

input
100 100
98 37
91 37
60 92
46 27
...

correct output
60

user output
73

Test 8

Group: 2

Verdict:

input
100 50
74 95
53 72
69 85
14 13
...

correct output
34

user output
76

Test 9

Group: 2

Verdict:

input
100 40
28 76
10 81
13 52
46 83
...

correct output
29

user output
79

Test 10

Group: 2

Verdict:

input
100 20
27 35
72 92
56 4
64 80
...

correct output
18

user output
81

Test 11

Group: 3

Verdict:

input
100000 200000
89244 59358
22943 56710
63331 89437
56581 38400
...

correct output
102510

user output
107130

Test 12

Group: 3

Verdict:

input
100000 100000
21701 85599
61542 21474
38081 29362
46316 64038
...

correct output
60593

user output
79082

Test 13

Group: 3

Verdict:

input
100000 50000
86469 4833
16351 35505
59315 33011
95464 16985
...

correct output
35933

user output
76382

Test 14

Group: 3

Verdict:

input
100000 40000
5392 23534
63204 45619
74330 25925
59678 88427
...

correct output
30074

user output
77922

Test 15

Group: 3

Verdict:

input
100000 20000
80156 16531
71753 77661
7028 33389
17168 646
...

correct output
16882

user output
85037