补提报告...

图着色问题 (25分)

这题需要注意的点是，题目中说无向图的颜色书为k个，那么这个图的颜色多了少了都是不对的。

#include <bits/stdc++.h>

using namespace std;
const int N = 550;

int n, m, k, t, v[N], e[N][N];
bool vis[N];

int main()
{
    cin >> n >> m >> k;
    while(m --){
        int x, y;
        cin >> x >> y;
        e[x][y] = 1;
        e[y][x] = 1;
    }
    cin >> t;
    while(t --){
        
        memset(vis, 0, sizeof(vis));
        int cnt = 0;
        for(int i = 1;i <= n; ++ i){
            int x;
            cin >> x;
            v[i] = x;
            if(!vis[x])
            vis[x] = true, cnt ++;
        }
        int f = 0;
        for(int i = 1;i <= n; ++ i){
            for(int j = 1;j <= n; ++ j){
                if(e[i][j] && v[i] == v[j])
                 f = 1;
            }
        }
        if(cnt != k) f = 1;
        if(!f) puts("Yes");
        else puts("No");
    }
}

愿天下有情人都是失散多年的兄妹 (25分)

需要注意的是，他的父母也要分出来男女。(毕竟可以离婚嘛...)

#include <bits/stdc++.h>

using namespace std;
const int N = 100010;

struct node{
    char f;
    int p1, p2;
}a[N];
int n, m;
vector<int> s1, s2;

void dfs(vector<int>& s, int x, int y, int deep){
    if(deep > 4) return ;
    if(x == -1 && y == -1) return ;

    if(x != -1)
    s.push_back(x);

    if(y != -1)
    s.push_back(y);
    
    int x1, y1, x2, y2;
    
    if(x == -1){
        x1 = -1;
        y1 = -1;
    }
    
    else x1 = a[x].p1, y1 = a[x].p2;
    
    if(y == -1){
        x2 = -1;
        y2 = -1;
    }
    else x2 = a[y].p1, y2 = a[y].p2;
    
    dfs(s, x1, y1, deep + 1);
    dfs(s, x2, y2, deep + 1);
}

bool check(int x, int y){
    s1.clear(), s2.clear();
    dfs(s1, a[x].p1, a[x].p2, 1);
    dfs(s2, a[y].p1, a[y].p2, 1);
    for(int i = 0;i < s1.size(); ++ i){
        for(int j = 0;j < s2.size(); ++ j){
            if(s1[i] == s2[j]) return false;
        }
    }
    
    return true;
}
int main()
{
    cin >> n;
    for(int i = 1;i <= n; ++ i){
        int x, id, p1, p2;
        char ch;
        cin >> id >>ch >> p1 >> p2;
        a[id].f = ch;
        a[id].p1 = p1;
        a[id].p2 = p2;
        a[p1].f = 'M';
        a[p2].f = 'F';
        
        if(!a[p1].p1) a[p1].p1 = -1;
        if(!a[p1].p2) a[p1].p2 = -1;
        if(!a[p2].p1) a[p2].p1 = -1;
        if(!a[p2].p2) a[p2].p2 = -1;
    }    
    cin >> m;
    while(m --){
        int x, y;
        cin >> x >> y;
        if(a[x].f == a[y].f){
            puts("Never Mind");
            continue;
        }
        check(x, y) == true ? puts("Yes") : puts("No");
    }
} 

肿瘤诊断 (30分)

三个方向的bfs

#include <bits/stdc++.h>

using namespace std;

struct node{
    int x, y, z;
};
int m, n, l, t, a[80][1450][150];
int tx[10] = {0, 0, -1, 1, 0, 0};
int ty[10] = {1, -1, 0, 0, 0, 0};
int tz[10] = {0, 0, 0, 0, 1, -1};
bool vis[80][1450][150];

int bfs(int x, int y, int z)
{
    int cnt = 0;
    vis[x][y][z] = true;
    queue<node> q;
    q.push({x, y, z});
    cnt ++;
    while(q.size())
    {
        node now = q.front();
        q.pop();
        for(int i = 0;i < 6; ++ i)
        {
            int nx = now.x + tx[i], ny = now.y + ty[i], nz = now.z + tz[i]; 
            if(!vis[nx][ny][nz] && a[nx][ny][nz])
            q.push({nx, ny, nz}), vis[nx][ny][nz] = true, cnt ++; 
        }
    }
    return cnt;
}
int main()
{
    cin >> m >> n >> l >> t;
    for(int i = 1;i <= l; ++ i)
        for(int j = 1;j <= m; ++ j)
            for(int k = 1;k <= n; ++ k)
                cin >> a[i][j][k];
    int res = 0;            
    for(int i = 1;i <= l; ++ i)
        for(int j = 1;j <= m; ++ j)
            for(int k = 1;k <= n; ++ k){
                if(a[i][j][k] && !vis[i][j][k]){
                    int tmp = bfs(i, j, k);
                    if(tmp >= t) res += tmp;
                }    
            }
    cout << res << endl;        
} 

红色警报 (25分)

每删除一个点判断一次是否比前面的“区域”多

#include <bits/stdc++.h>
//代码看师哥的，自己搜不明白QAQ 
using namespace std;
const int N = 550;

int n, m, k, a[N][N];
bool vis[N], e[N];

void dfs(int s)
{
    vis[s] = true;
    for(int i = 0;i < n; ++ i){
        if(a[s][i] && !vis[i])
            dfs(i);
    }
}

int get_sum()
{
    int cnt = 0;
    memset(vis, 0, sizeof(vis));
    for(int i = 0;i < n; ++ i)
    {
        if(!vis[i] && !e[i])
            dfs(i), cnt ++;
    }
    return cnt;
}

int main()
{
    cin >> n >> m;
    while(m --)
    {
        int x, y;
        cin >> x >> y;
        a[x][y] = 1;
        a[y][x] = 1;
    }
    cin >> k;
    int cnt1 = get_sum(), cnt2, cnt = 0;
    while(k --)
    {
        int x;
        cin >> x;
        e[x] = true;
        for(int i = 0;i < n; ++ i)
         if(a[x][i]) 
            a[x][i] = a[i][x] = 0;
        
        cnt2 = get_sum();
        if(cnt1 >= cnt2)
        printf("City %d is lost.\n", x);
        else printf("Red Alert: City %d is lost!\n", x);
        if(++cnt == n) puts("Game Over.");
        cnt1 = cnt2;
    }
}