洛谷 6道简单图论

最新推荐文章于 2024-06-30 01:12:36 发布

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原文链接：http://www.cnblogs.com/CzYoL/p/7767442.html

本文精选了洛谷平台上的经典算法题目，包括消息扩散、灾后重建等，详细介绍了Tarjan缩点、Floyd算法、最短路径、二分查找、Dijkstra算法等核心算法的应用场景与实现细节。

luogu 2002 消息扩散

tarjan缩点:对原图进行tarjan缩点，对缩后的点连边，统计入度为0的点。

luogu 1119 灾后重建

Floyed:对于每个新加入的点，将其作为中转点，跑floyed。

luogu 1186 玛丽卡

最短路：类似次小生成树，求出最短路径后，将其路径上的每一条边删去，再求一遍。

luogu 1462 通往奥格瑞玛的道路

最短路+二分:二分最大值，将值大于二分值的都删去，看s和t是否连通。

luogu 1613 跑路

floyed+倍增:首先类似传递闭包或倍增求出两点间是否存在一条\(2^k\)的道路，若存在初始化dis[i][j] = 1,然后跑floyed。

luogu 2700 逐个击破

最大生成树:将边从大至小排序，将每个点看作一个集合，先将k个特殊点染色，然后从大至小枚举边，将边两边的点集合分情况合并：

一端染色，一端未染色：未染色端染上色，两集合合并。
两端都未染色：两集合直接合并。
两端都染色：加上该边的权值，合并集合。

code

luogu 2002

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 50, M = 5e5 + 50;
int n, m, low[N], dfn[N], clk, sccno[N], sccnum[N], sccCnt, ans, outdeg[N], indeg[N], from[N];
vector<int> G[N];
stack<int> S;

inline void dfs(int u){
    dfn[u] = low[u] = ++clk;
    S.push(u);

    for(int e = G[u].size() - 1; e >= 0; e--){
        int v = G[u][e];
        if(!dfn[v]){
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v]) low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u]){
        int x; sccCnt++;
        for(;;){
            x = S.top(); S.pop();
            sccno[x] = sccCnt;
            sccnum[sccCnt]++;
            if(x == u) return;
        }
    }
}

int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1; i<= m; i++){
        int x, y; scanf("%d%d", &x, &y);
        if(x != y) G[x].push_back(y);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) dfs(i);
//    for(int i = 1; i <= n; i++) cout<<sccno[i]<<" "<<endl;
    for(int i = 1; i <= n; i++){
        for(int e = G[i].size() - 1; e >= 0; e--){
            int v = G[i][e];
            if(sccno[i] != sccno[v])
                outdeg[sccno[i]]++, indeg[sccno[v]]++;
        }
    }
    for(int i = 1; i <= sccCnt; i++){
        if(!indeg[i] && !outdeg[i]) ans++;
        else if(!indeg[i]) ans ++;
    }
    printf("%d", ans);
    return 0;
}

luogu 1119

#include<bits/stdc++.h>
using namespace std;
const int N = 220, OO = 0x3f3f3f3f;
int n, m;
int f[N][N], q;
vector<int> h[1000000];
bool vst[N];

int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1 ; i <= n; i++){
        int x; scanf("%d", &x);
        h[x].push_back(i);
    }

    memset(f, OO, sizeof f);
    for(int i = 1; i <= n; i++) f[i][i] = 0;

    for(int i = 1; i <= m; i++){
        int u, v, w; scanf("%d%d%d", &u, &v, &w);
        u++, v++;
        f[u][v] = f[v][u] = min(f[u][v], w);
    }


    scanf("%d", &q);
    int now = -1;
    for(int i = 1; i <= q; i++){
        int x, y, t; scanf("%d%d%d", &x, &y, &t);
        x++, y++;
        while(now < t){
            now++;
            for(int j = 0; j < h[now].size(); j++){
                int u = h[now][j];
//                cout<<now<<" "<<u<<endl;
                vst[u] = true;
                for(int k = 1; k <= n; k++)
                    for(int l = 1; l <= n; l++)
                        f[k][l] = min(f[k][l], f[k][u] + f[u][l]);
            }
        }
        printf("%d\n", (f[x][y] < OO && vst[x] && vst[y]) ? f[x][y] : -1);
    }
    return 0;
}

luogu 1186

#include<bits/stdc++.h>
using namespace std;
const int N = 1005, M = 1e6 + 5, OO = 0x3f3f3f3f;
int n, ans, tmp, m;
vector<pair<int, int> > G[N];
vector<pair<int, int> > route;
pair<int, int> from[N];
int dis[N];
#define mp make_pair
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > que;

inline void dj(int s, int t, int f, int a, int b){
    while(!que.empty()) que.pop();
    memset(dis, OO, sizeof dis);
    dis[s] = 0;
    que.push(mp(0, s));
    while(!que.empty()){
        pair<int, int> tt = que.top(); que.pop();
        int u = tt.second, dist = tt.first;
//        cout<<u<<"->";
        if(u == t) break;
        for(int e = G[u].size() - 1; e >= 0; e--){
            int v = G[u][e].first;
            int len = G[u][e].second;
            if(f == 1 && ((u == a && v == b)|| (u == b && v == a))) continue;
//            cout<<v<<" !!";
            if(dis[v] > dis[u] + len){
                from[v] = mp(u, len);
                dis[v] = dis[u] + len;
                que.push(mp(dis[v], v));
            }
        }
//        cout<<endl;
    }
}

int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++){
        int a, b, c; scanf("%d%d%d", &a ,&b, &c);
        G[a].push_back(mp(b, c));
        G[b].push_back(mp(a, c));
    }
    dj(1, n, 0, 0, 0);
    ans = tmp = dis[n];
//    cout<<dis[n]<<endl;
    int now = n;
    route.push_back(mp(now, 0));
    while(from[now].first) route.push_back(from[now]), now = from[now].first;
//    for(int i = 0; i <= route.size() - 1; i++) cout<<route[i].first<<" "<<route[i].second<<endl;
    for(int i = 0; i < route.size() - 1; i++){
        int u = route[i].first, v = route[i + 1].first;
        dj(1, n, 1, u, v);
//        cout<<u<<" "<<v<<" "<<dis[v]<<endl;
        ans = max(ans, dis[n]);
    }
    printf("%d", ans);
}

luogu 1462

#include<bits/stdc++.h>
using namespace std;
const int N = 10005, M = 1e6 + 5;
const long long OO = 2e18;
int n, m, b;
long long maxx = -OO, minn = OO, f[N];
vector<pair<int, long long> > G[N];
long long dis[N];
#define mp make_pair
priority_queue<pair<long long, int>, vector<pair<long long, int> >, greater<pair<long long, int> > > que;

inline bool check(long long mid){
    if(f[1] > mid) return false;
    while(!que.empty()) que.pop();
    for(int i = 0; i <= n; i++) dis[i] = OO;
    dis[1] = 0;
    que.push(mp(0, 1));
    while(!que.empty()){
        pair<long long, int> t = que.top(); que.pop();
        if(t.second == n) return t.first <= b;
        int u = t.second;
        for(int e = G[u].size() - 1; e >= 0; e--){
            int v = G[u][e].first, len = G[u][e].second;
            if(f[v] > mid) continue;
            if(dis[v] > dis[u] + len){
                dis[v] = dis[u] + len;
                que.push(mp(dis[v], v));
            }
        }
    }
    return false;
}

int main(){
    scanf("%d%d%d", &n, &m, &b);
    for(int i = 1; i <= n; i++) scanf("%d", &f[i]), maxx = max(maxx, f[i]), minn = min(minn, f[i]);
    for(int i = 1; i <= m; i++){
        int x, y, c; scanf("%d%d%d", &x, &y, &c);
        G[x].push_back(mp(y, c)), G[y].push_back(mp(x, c));
    }
    long long l = minn, r = maxx, ans = -1;
    while(l <= r){
        long long mid = l + r >> 1;
        if(check(mid)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    if(ans == -1) printf("AFK");
    else printf("%lld", ans);
    return 0;
}

luogu1613

#include<bits/stdc++.h>
using namespace std;
const int N = 55, OO = 0x3f3f3f3f;
int n, dis[N][N], m;
bool f[N][N][35];

int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++){
        int x, y; scanf("%d%d", &x, &y);
        f[x][y][0] = true;
    }
    memset(dis, OO, sizeof dis);
    for(int l = 0; l <= 31; l++)
        for(int k = 1; k <= n; k++)
            for(int i = 1; i <= n; i++)
                for(int j = 1; j <= n; j++){
                        if(l) f[i][j][l]= f[i][j][l] || (f[i][k][l - 1] && f[k][j][l - 1]);
                        if(f[i][j][l]) dis[i][j] = 1;
                    }
//  for(int i = 1; i <= n; i++)
//          for(int j = 1; j <= n; j++)
//              for(int l = 1; l <= 5; l++)
//                  cout<<i<<" "<<j<<" "<<l<<" "<<f[i][j][l]<<endl;
//  for(int i = 1; i <= n; i++)
//      for(int j = 1; j <= n; j++)
//          cout<<i<<" "<<j<<" "<<dis[i][j]<<endl;
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
    printf("%d", dis[1][n]);
    return 0;
}

luogu2700

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+50;
int n, k, anc[N];
long long ans;
bool mark[N];
struct node{
    int x, y, c;
    inline bool operator < (const node &b) const{
        return c > b.c;
    }
}edge[N];

inline int getAnc(int x){
    return x == anc[x] ? x : (anc[x] = getAnc(anc[x]));
}

int main(){
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) anc[i] = i;
    for(int i = 1; i <= k; i++){
        int x; scanf("%d", &x);
        x++;
        mark[x] = true;
    }
    for(int i = 1; i < n; i++){
        int x, y, c; 
        scanf("%d%d%d", &x, &y, &c);
        x++, y++;
        edge[i] = (node){x, y, c};
    }
    sort(edge + 1, edge + n);
    for(int i = 1; i < n; i++){
        int x = edge[i].x, y = edge[i].y, c = edge[i].c;
        int fx = getAnc(x), fy = getAnc(y);
        if(fx != fy){
            if(mark[fx] && !mark[fy]){
                mark[fy] = true;
                anc[fx] = fy;
            }
            else if(!mark[fx] && mark[fy])
                anc[fx] = fy;
            else if(mark[fx] && mark[fy]){
                anc[fx] = fy;
                ans += 1ll*c;
            }
            else if(!mark[fx] && !mark[fy])
                anc[fx] = fy;
        }
//      cout<<x<<" "<<y<<" "<<c<<" "<<ans<<endl;
    }
    printf("%lld", ans);
    return 0;
}

转载于:https://www.cnblogs.com/CzYoL/p/7767442.html