BZOJ4945 NOI2017 游戏

这题放在NOI里应该不算难的吧……但是细节比较多，所以写起来会有点**

题目限定了道路不能通行某种车辆，也就是可以通行两种车辆

我们将这两种车辆分别作为正点和反点进行约束就可以了

建图较为容易

最后将所有的x枚举一下即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cstdlib>

using namespace std;

vector <int> g[100005];
vector <int> sta;
vector <int> pos;
int compcnt = 0;
vector <int> comp[100005];

int dfn[100005], low[100005], is[100005], ic[100005], idx, cnt, n, m, indeg[100005], ans;

int seq[100005], lim[100005][5], d;

char str[100005];

void tarjan(int u) {
    is[u] = 2;
    low[u] = dfn[u] = ++idx;
    sta.push_back(u);
    for (int i = 0; i < g[u].size(); i++) {
        if (is[g[u][i]] == 0) {
            tarjan(g[u][i]);
            low[u] = min(low[u], low[g[u][i]]);
        }
        else {
            if (is[g[u][i]] == 2) {
                low[u] = min(low[u], dfn[g[u][i]]);
            }
        }
    }
    if (low[u] == dfn[u]) {
        ++cnt;
        for (int j = 0; j != u;) {
            j = sta[sta.size() - 1];
            sta.pop_back();
            is[j] = 1;
            ic[j] = cnt;
            comp[cnt].push_back(j);
        }
    }
}

bool check() {
    for(int i=2;i<=2*n+2;i+=2) {
        if(ic[i]==ic[i-1]&&ic[i]) {
            return false;
        }
    }
    return true;
}

inline void link(int p,int q){
    
    g[p].push_back(q);
}

int frow(int typ,int key){
    if(typ==1) return key==2;
    if(typ==2) return key==1;
    if(typ==3) return key==1;
}

int main()
{
    scanf("%d%d",&n,&d);
    
    scanf("%s",&str);
    
    for (int i=1;i<=n;i++) seq[i]=(str[i-1]=='x'?0:(str[i-1]-'a'+1));
    for (int i=1;i<=n;i++) if(seq[i]==0) pos.push_back(i);
    
    scanf("%d", &m);
    
    char c1[2],c2[2];
    
    for (int i=1;i<=m;i++) {
        scanf("%d%s%d%s",&lim[i][0],&c1,&lim[i][2],&c2);
        lim[i][1]=c1[0]-'A'+1;
        lim[i][3]=c2[0]-'A'+1;

    }
    
    for (int pix=0;pix<1<<d;pix++) {
        memset(dfn,0,sizeof dfn); memset(low,0,sizeof low); memset(is,0,sizeof is); memset(ic,0,sizeof ic); 
        sta.clear(); 
        for(int i=1;i<=cnt;i++) comp[i].clear();
        idx=0; cnt=0;

        int tmp = pix;
        
        for (int i=0;i<2*n+2;i++) g[i].clear();
 
        for (int j=pos.size()-1;j>=0;j--) {
            seq[pos[j]]=tmp%2 + 1;
            tmp/=2;
        }
        
        for (int i=1;i<=m;i++) { 
            if (lim[i][1]==seq[lim[i][0]]) continue;
            if (lim[i][3]==seq[lim[i][2]]) link(2*lim[i][0]-frow(seq[lim[i][0]],lim[i][1]),2*lim[i][0]-1+frow(seq[lim[i][0]],lim[i][1]));
            else link(2*lim[i][0]-frow(seq[lim[i][0]],lim[i][1])
                     ,2*lim[i][2]-frow(seq[lim[i][2]],lim[i][3])),
                      link(2*lim[i][2]-1+frow(seq[lim[i][2]],lim[i][3]),
                           2*lim[i][0]-1+frow(seq[lim[i][0]],lim[i][1]));    

        }
        for (int i = 1; i <= 2*n; i++) {
            if (dfn[i] == 0)
                tarjan(i);
        }
        
        if(!check()) continue;
        for (int i=1;i<=2*n;i+=2) {
            if(ic[i]<ic[i+1]) 
                printf("%c",'A'-'a'+(seq[i/2+1]==1?'b':'a'));    
            else printf("%c",'A'-'a'+(seq[i/2+1]==3?'b':'c'));
        }
        printf("
");
        return 0;
    }
    printf("-1
");
    
    return 0;
}