标签：ios   type   return   signed   att   break   space   res   bool   

题意：

在一个\(100*100\)的方格中，要求从\(b\)走到\(g\)，途中经过\(c\)但不经过\(u\)，并且不能走已经做过的路。如果可以，就求出路径。

思路：

拆点建费用流，看能不能从\(c\)走两条路走到\(b，g\)。然后输出路径。

代码：

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<bitset>
#include<string>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 20000 + 5;
const int M = 50 + 5;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const ll MOD = 1000000007;

struct Edge{
    int to, next, cap, cost;
}edge[10000 * 4 * 10 + 100];
int head[maxn], tot;
int pre[maxn], dis[maxn];
bool vis[maxn];
int N;
void init(){
    N = maxn;
    tot = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int cap, int cost){  //双向边
    edge[tot].to = v;
    edge[tot].cap = cap;
    edge[tot].cost = cost;
    edge[tot].next = head[u];
    head[u] = tot++;

    edge[tot].to = u;
    edge[tot].cap = 0;
    edge[tot].cost = -cost;
    edge[tot].next = head[v];
    head[v] = tot++;
}
bool spfa(int s, int t){
    queue<int> q;
    for(int i = 0; i <= N; i++){
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u];i != -1;i = edge[i].next){
            int v = edge[i].to;
            int c = edge[i].cap;
            int w = edge[i].cost;
            if(c && dis[v] > dis[u] + w){
                dis[v] = dis[u] + w;
                pre[v] = i;
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}
int MCMF(int s, int t, int &cost){
    int flow = 0;
    cost = 0;
    while(spfa(s, t)){
        int MIN = INF;
        for(int i = pre[t]; i != -1;i = pre[edge[i ^ 1].to]){
            MIN = min(MIN, edge[i].cap);
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i ^ 1]. to]){
            edge[i]. cap -= MIN;
            edge[i ^ 1]. cap += MIN;
            cost += edge[i]. cost * MIN;
        }
        flow += MIN;
    }
    return flow;
}

int n, m ,bx, by, cx, cy, gx, gy, ux, uy;
int getid(int x, int y, int out){
    return (x - 1) * m + y + n * m * out;
}
char getturn(int a, int b){
    int x1 = (a - 1) / m + 1, x2 = (b - 1) / m + 1;
    int y1 = a - (x1 - 1) * m, y2 = b - (x2 - 1) * m;
    if(x1 + 1 == x2) return 'U';
    if(x1 - 1 == x2) return 'D';
    if(y1 + 1 == y2) return 'R';
    if(y1 - 1 == y2) return 'L';
}

/***********************
输出路径
string step;
int ans[20000], cnt;
vector<int> g[maxn];
void dfs(int s, int t){
    vis[s] = 1;
    if(s == t) return;
    for(int i = 0; i < g[s].size(); i++){
        int v = g[s][i];
        if(!vis[v]){
            vis[v] = 1;
            ans[cnt++] = v;
            dfs(v, t);
            break;
        }
    }
}
void path(int st, int en){
    for(int i = 1; i <= n * m; i++){
        g[i].clear();
        for(int j = head[i + n * m]; j != -1; j = edge[j].next){
            int c = edge[j].cap;
            int v = edge[j].to;
            if(!c && i != v){   //流量为0
                g[i].push_back(v);
            }
        }
    }
    string tmp;

*************************/

    memset(vis, 0, sizeof(vis));
    step = "";
    cnt = 0;
    ans[cnt++] = getid(cx, cy, 0);
    dfs(st, en), cnt--;
    tmp = "";
    if(ans[cnt - 1] == getid(gx, gy, 0)){
        tmp = "";
        for(int i = 0; i < cnt - 1; i++){
            tmp += getturn(ans[i], ans[i + 1]);
        }
        step = step + tmp;
    }
    else{
        tmp = "";
        for(int i = cnt - 1; i > 0; i--){
            tmp += getturn(ans[i], ans[i - 1]);
        }
        step = tmp + step;
    }

    cnt = 0;
    ans[cnt++] = getid(cx, cy, 0);
    dfs(st, en), cnt;
    tmp = "";
    if(ans[cnt - 1] == getid(gx, gy, 0)){
        tmp = "";
        for(int i = 0; i < cnt - 1; i++){
            tmp += getturn(ans[i], ans[i + 1]);
        }
        step = step + tmp;
    }
    else{
        tmp = "";
        for(int i = cnt - 1; i > 0; i--){
            tmp += getturn(ans[i], ans[i - 1]);
        }
        step = tmp + step;
    }
}
int main(){
    int cost;
    int to[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
    while(scanf("%d%d", &n, &m) && n + m){
        init();
        scanf("%d%d", &bx, &by);
        scanf("%d%d", &cx, &cy);
        scanf("%d%d", &gx, &gy);
        scanf("%d%d", &ux, &uy);
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                int x, y;
                addEdge(getid(i, j, 0), getid(i, j, 1), 1, 1);
                addEdge(getid(i, j, 1), getid(i, j, 0), 1, 1);
                for(int k = 0; k < 4; k++){
                    x = i + to[k][0];
                    y = j + to[k][1];
                    if(x < 1 || y < 1 || x > n || y > m) continue;
                    if((x == ux && y == uy) || (i == ux && j == uy)) continue;
                    addEdge(getid(i, j, 1), getid(x, y, 0), 1, 1);
                }
            }
        }
        int st = getid(cx, cy, 1), en = 2 * n * m + 1;
        addEdge(getid(bx, by, 1), en, 1, 1);
        addEdge(getid(gx, gy, 1), en, 1, 1);
        int flow = MCMF(st, en, cost);
        if(flow < 2) printf("NO\n");
        else{
            path(getid(cx, cy, 0), en);
            printf("YES\n");
            cout << step << endl;
        }
    }
    return 0;
}

Kattis amazingadventures Amazing Adventures（费用流路径）题解

标签：ios   type   return   signed   att   break   space   res   bool   

原文地址：https://www.cnblogs.com/KirinSB/p/11609532.html