暑假第十一测

今天爆零了，roadblock是原题，结果还审错题了。。。

 

题解

第一题：把最短路上的边提出来，然后跑最大流直到跑不动；但这样写比较麻烦；

用最小费用最大流也可以过，比某些网络流还快，这东西也玄学；

#include <bits/stdc++.h>

using namespace std;
const int maxn = 3010, maxm = 41000, inf = 1000000008;
struct Netflow{
    int tot,head[maxn],maxflow,S,T,minlen;
    int inq[maxn],pre[maxn],pree[maxn],dis[maxn];
    struct edge{
        int to,f,w,nxt;
    }G[maxm];
    void init(int s, int t){
        S = s, T = t;
        tot = 1;
        memset(head,0,sizeof(head));
        maxflow = 0;
        minlen = inf;
    }
    void add(int u,int v,int f,int w){
        G[++tot].nxt = head[u];
        G[tot].to = v;
        G[tot].f = f;
        G[tot].w = w;
        head[u] = tot;
        G[++tot].nxt = head[v];
        G[tot].to = u;
        G[tot].f = 0;
        G[tot].w = -w;
        head[v] = tot;
    }
    void SPFA(){
        queue <int> Q;
        for(int i = 0; i <= T; i++)dis[i] = inf;
        memset(inq,0,sizeof(inq));
        Q.push(S);
        inq[S] = 1;
        dis[S] = 0;

        while(!Q.empty()){
            int u = Q.front();
            Q.pop();
            inq[u] = 0;

            for(int i = head[u]; i; i= G[i].nxt){
                int v = G[i].to;
                if(G[i].f && dis[v] > dis[u] + G[i].w){
                    dis[v] = G[i].w + dis[u];
                    if(!inq[v]){
                        Q.push(v);
                        inq[v] = 1;
                    }
                }
            }
        }
        minlen = dis[T];
    }
    bool spfa(){
        queue <int> Q;
        for(int i = 0; i <= T; i++)dis[i] = inf;
        memset(inq,0,sizeof(inq));
       // memset(pre,0,sizeof(pre));
       // memset(pree,0,sizeof(pree));

        Q.push(S);
        inq[S] = 1;
        dis[S] = 0;

        while(!Q.empty()){
            int u = Q.front();
            Q.pop();
            inq[u] = 0;

            for(int i = head[u]; i; i= G[i].nxt){
                int v = G[i].to;
                if(G[i].f && dis[v] > dis[u] + G[i].w){
                    dis[v] = G[i].w + dis[u];
                    pre[v] = u;
                    pree[v] = i;
                    if(!inq[v]){
                        Q.push(v);
                        inq[v] = 1;
                    }
                }
            }
        }
        return dis[T] == minlen;
    }
    void augment(){
        while(spfa()){
            int u = T, delta = inf;
            while(u != S){
                delta = min(delta, G[pree[u]].f);
                u = pre[u];
            }
            u = T;
            while(u != S){
                G[pree[u]].f -= delta;
                G[pree[u]^1].f += delta;
                u = pre[u];
            }
            maxflow += delta;
        }
        printf("%d
",maxflow);
    }
};


int main()
{
    freopen("roadblock.in","r",stdin);
    freopen("roadblock.out","w",stdout);
    int T,n,m;
    scanf("%d",&T);
    while(T--){
        Netflow Tr;
        scanf("%d%d",&n,&m);
        Tr.init(1,n);
        while(m--){
            int u, v, w;
            scanf("%d%d%d",&u,&v,&w);
            Tr.add(u, v, w, 1);
            Tr.add(v, u, w, 1);
        }
        Tr.SPFA();
        Tr.augment();
    }

    return 0;
}
View Code

第二题：

一道拓扑排序，我完全没看出来，在那里傻逼并查集了半天；

可以知道一条路径中间忽略的车站等级一定更低，但是暴力建边复杂度是(m*n)不能过；

所以就有了传说已久的线段树优化建边，学std写了一波，线段树从下往上建边，对于出现的一组要求我们就开一个新节点；

至于无向图的线段树建边还要稍稍麻烦些，要开两棵树，有空看看吧

#include<bits/stdc++.h>
using namespace std;

const int M = 600005;
int in[M], h[M], mark[M], dg[M], tot, idx, n, m, pos[M];
struct edge{
    int v, nxt;
}G[M * 10];
void add(int u, int v){
    G[++tot].nxt = h[u]; G[tot].v = v; h[u] = tot; in[v]++;
}

struct Node {
    Node *ls , *rs;
    int id;
}pool[M << 1], *root, *tail = pool;

Node * build(int l = 1, int r = n){
    Node * nd = ++tail;  nd->id = ++idx; 
    if(l == r) mark[idx] = 1, pos[l] = idx;
    else {
        int mid = (l + r) >> 1;
        nd->ls = build(l, mid);
        add(nd->ls->id, nd->id);
        nd->rs = build(mid + 1, r);
        add(nd->rs->id, nd->id);
    }
    return nd;
}
#define Ls nd->ls, l, mid
#define Rs nd->rs, mid+1, r
void modify(int L, int R, int crtime, Node * nd = root, int l = 1, int r = n){
    if(L <= l && r <= R){
        add(nd->id, crtime);
    }
    else{
        int mid = (l + r) >> 1;
        if(L <= mid)modify(L, R, crtime, Ls);
        if(R > mid)modify(L, R, crtime, Rs);
    }
}
void solve(){
    int ans = 1;
    queue <int> q;
    q.push(0);
    for(int i = 1; i <= n; i++)
        if(!in[pos[i]]) q.push(pos[i]), dg[pos[i]] = 1;
    while(!q.empty()){
        int u = q.front(); q.pop();
        for(int i = h[u]; i; i = G[i].nxt){
            int v = G[i].v;
            in[v]--;
            dg[v] = max(dg[v], dg[u] + mark[v]);
            if(!in[v])q.push(v);
            ans = max(ans, dg[v]);
        }
        //for(int i = 1; i <= n; i++)printf("%d ", dg[pos[i]]);puts("");
    }
    printf("%d
", ans);
}

int main(){
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
    scanf("%d%d", &n, &m);
    root = build();
    int siz, lst, now;
    for(int i = 1; i <= m; i++){
        scanf("%d%d", &siz, &lst);
        add(0, ++idx);
        add(idx, pos[lst]);
        for(int j = 2; j <= siz; j++){
            scanf("%d", &now);
            add(idx, pos[now]);
            modify(lst+1, now-1, idx);
            lst = now;
        }
    }
    solve();
}
View Code

第三题：一道巨复杂的费用流，又颓了没听，以下是std

//2s 512M
#define PN "trouble"
#include <cstdio>
#include <cstring>
#include <algorithm>

template<class T>inline void readin(T &res) {
    static char ch;T flag = 1;
    while((ch=getchar())<'0'||ch>'9')if(ch=='-')flag=-1;
    res=ch-48;
    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-48;
    res*=flag;
}
template<class T>inline T abs(T a) {return a >= 0? a: -a;}
template<class T>inline T max(T a,T b) {return a >= b? a: b;}
template<class T>inline T min(T a,T b) {return a <= b? a: b;}

const int N = 40 + 10;
const int INF = 0x3f3f3f3f;
const int NODES = 10 * N * N;
const int EDGES = 100 * N * N;
struct EDGE {int v, res, cost, upre;} g[EDGES];
int head[NODES], ne = 1;
inline void adde(int u,int v,int cap,int cost) {
    g[++ne] = (EDGE){v,cap,cost,head[u]};head[u] = ne;
    g[++ne] = (EDGE){u,0,-cost,head[v]};head[v] = ne;
}

int s, t;
int flow, cost;

#include <queue>
std::queue<int> q;
int d[NODES], a[NODES], p[NODES];
bool inq[NODES];
bool SPFA() {
    while(!q.empty()) q.pop();
    memset(d,63,sizeof(int)*(t+1));
    memset(inq,false,sizeof(int)*(t+1));
    d[s] = 0;inq[s] = true;q.push(s);
    a[s] = INF;

    int u, i, v;
    while(!q.empty()) {
        u = q.front();q.pop();inq[u] = false;
        for( i = head[u]; i; i = g[i].upre ) if(g[i].res && d[v = g[i].v] > d[u] + g[i].cost) {
            d[v] = d[u] + g[i].cost;
            if(!inq[v]) inq[v] = true, q.push(v);
            p[v] = i;a[v] = min(a[u],g[i].res);
        }
    }
    if(d[t] == INF) return false;
    flow += a[t];cost += a[t] * d[t];
    for( u = t; u != s; g[p[u]].res -= a[t], g[p[u]^1].res += a[t], u = g[p[u]^1].v );
    return true;
}

int T, n, m, A, B, t2, Q;
bool Ball[N][N];
inline int HOLE(int x,int y,int z) {return 3 * (x * m + y) + z + 1;}
int main() {
    freopen(PN".in","r",stdin);
    freopen(PN".out","w",stdout);
    readin(T);//T=8~12 ans>0 print 1
    readin(n);readin(m);readin(A);readin(B);

    s = 0;t = 3 * n * m + 2;
    memset(head,0,sizeof(int)*(t+1));ne = 1;

    t2 = 3 * n * m + 1;adde(t2,t,0,0);
    for( int i = 0, j; i < n; i++ ) for( j = 0; j < m; j++ ) {
        scanf("%1d",&Ball[i][j]);Ball[i][j] ^= 1;
        if((i+j)&1) {
            adde(s,HOLE(i,j,0),1,0);
            adde(s,HOLE(i,j,0),1,A);
            adde(s,HOLE(i,j,0),1,2*A);
            adde(s,HOLE(i,j,0),1,3*A);
            adde(HOLE(i,j,0),HOLE(i,j,1),1,0);
            adde(HOLE(i,j,0),HOLE(i,j,1),1,B-A);
            adde(HOLE(i,j,0),HOLE(i,j,2),1,0);
            adde(HOLE(i,j,0),HOLE(i,j,2),1,B-A);
        } else {
            adde(HOLE(i,j,0),t2,1,0);
            adde(HOLE(i,j,0),t2,1,A);
            adde(HOLE(i,j,0),t2,1,2*A);
            adde(HOLE(i,j,0),t2,1,3*A);
            adde(HOLE(i,j,1),HOLE(i,j,0),1,0);
            adde(HOLE(i,j,1),HOLE(i,j,0),1,B-A);
            adde(HOLE(i,j,2),HOLE(i,j,0),1,0);
            adde(HOLE(i,j,2),HOLE(i,j,0),1,B-A);
        }
    }
    for( int i = 0, j; i < n; i++ ) for( j = 0; j < m; j++ )
        if(((i+j)&1) && Ball[i][j]) {
            if(j != 0 && Ball[i][j-1])
                adde(HOLE(i,j,1),HOLE(i,j-1,1),1,0);
            if(j != m - 1 && Ball[i][j+1])
                adde(HOLE(i,j,1),HOLE(i,j+1,1),1,0);
            if(i != 0 && Ball[i-1][j])
                adde(HOLE(i,j,2),HOLE(i-1,j,2),1,0);
            if(i != n - 1 && Ball[i+1][j])
                adde(HOLE(i,j,2),HOLE(i+1,j,2),1,0);
        }

    readin(Q);
    for( int x = 1, i; x <= Q; x++ ) {
        for( i = head[t2]; i; i = g[i].upre )
            if(g[i].v == t) g[i].res++;
        while(SPFA());
        printf("%d
",8 <= T && T <= 12?cost > 0:cost);
    }
    return 0;
}
View Code
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原文地址：https://www.cnblogs.com/EdSheeran/p/9394250.html