2021 RoboCom 世界机器人开发者大赛-本科组（复赛）

#include
using namespace std;

int a[10], b[10], c[10];

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T;  cin>>T;
    while(T--){
        for(int i = 1; i >a[i];
        //-1
        int ok = 1;
        for(int i = 1; i >b[i];  c[i] = b[i]-a[i];
            if(c[i]%20!=0)ok = 0;
            c[i] /= 20;
        }
        if(a[1]+a[2]+a[3] != b[1]+b[2]+b[3])ok = 0;
        int mod = (c[1]%3+3)%3;
        for(int i = 2; i 0)+(c[2]>0)+(c[3]>0);
        if(num==2)c[1]=-c[1], c[2]=-c[2], c[3]=-c[3];
        int x = min({c[1],c[2],c[3]}), z = max({c[1],c[2],c[3]}), y = c[1]+c[2]+c[3]-x-z;//排序
        int res = 0;
        res += -y;  x -= y;  z += y*2;  y = 0; //操作1
        res += z/3*2;  //操作2
        cout

//15分
#include
using namespace std;
const int maxn = 1010;
int v[maxn], w[maxn];
int f[maxn*maxn];
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n, m;  cin>>n>>m;
    for(int i = 1; i >v[i];
    for(int i = 1; i >w[i];
    for(int i = 1; i = 0; j--){
            if(j>=v[i])f[j] = max(f[j], f[j-v[i]]+w[i]);
        }
    }
    cout

//25分
#include
using namespace std;
const int maxn = 1010;
int v[maxn], w[maxn];
int f[maxn*maxn];
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n, m;  cin>>n>>m;
    int sum = 0;
    for(int i = 1; i >v[i];
    for(int i = 1; i >w[i], sum += w[i];
    memset(f,0x3f,sizeof(f));
    f[0] = 0;
    for(int i = 1; i = w[i]; j--){
            f[j] = min(f[j], f[j-w[i]]+v[i]);
        }
    }
    for(int i = sum; i >= 0; i--){
        if(f[i]

#include
using namespace std;
typedef pair PII;
const int maxn = 1010;

int n, wmax, w1, w2;
vectorans;
int ans1, ans2;

dequeA, D;
queueB, C;
void update(){  //每次按DCBA的顺序处理一遍当前流水线上的箱子
    if(D.size()!=0){
        PII x = D.front();
        if(x.second == 0){ //x没装箱，给x装箱
            D.pop_front();
            D.push_front(make_pair(x.first,1));
        }else{
            D.pop_front();
            if(D.size()==0)D.push_front(x);//x不是最后一个就纳入出队考核
            else{
                PII y = D.front();
                if(y.second == 0){//y没装箱
                    if(y.first+x.first w2)ans2++; //x正式出队
                        else A.push_back(x);  //给x放回去A重生
                    }
                }else{
                    if(x.first > w2)ans2++;  //x正式出队
                    else A.push_back(x);     //给x放回去A重生
                }
                if(D.front().second == 0){ //z没装箱，给z装箱
                    PII z = D.front();  D.pop_front();
                    D.push_front(make_pair(z.first,1));
                }
            }
        }
    }
    if(C.size()!=0){
        PII x = C.front();  C.pop();
        if(x.first > w2)ans1++;
        else D.push_back(x);
    }
    if(B.size()!=0){
        PII x = B.front();  B.pop();
        if(x.first > w1)C.push(make_pair(x.first, 1));
        else C.push(x);
    }
    if(A.size()!=0){
        B.push(A.front());  A.pop_front();
    }
}

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>n>>wmax>>w1>>w2;
    for(int i = 1; i >x;  //把x放到A口,考虑是否装箱
        if(A.size()==0)A.push_back(make_pair(x,0));
        else{
            PII y = A.front();
            if(y.first + x D2; //临时队列
    while(D.size() != 0){
        PII x = D.front();  D.pop_front();
        if(x.first > w2)ans2++; else D2.push(x);
    }
    while(D2.size()!=0){ D.push_back(D2.front());  D2.pop(); }
    while(A.size()!=0){if(A.front().first != 0)ans.push_back(A.front().first); A.pop_front(); }
    while(B.size()!=0){if(B.front().first != 0)ans.push_back(B.front().first); B.pop(); }
    while(C.size()!=0){if(C.front().first != 0)ans.push_back(C.front().first); C.pop(); }
    while(D.size()!=0){if(D.front().first != 0)ans.push_back(D.front().first); D.pop_front(); }
    sort(ans.begin(), ans.end()); //递增排序
    cout

#include
using namespace std;
typedef pair PII;
typedef pair PIII;
const int maxn = 220, inf = 1e9+10;

int n, m, T, a[maxn][maxn];//地图和移动
int dx[] = {-1,1,0,0}, dy[] = {0,0,-1,1};
PII st;//大本营坐标
struct node{int x, y, cc;};
vectorjs; //僵尸坐标和血量
int cnt[maxn*maxn], now[maxn*maxn]; //每个僵尸剩余血量, 该僵尸走了第几步
vectorp[maxn*maxn]; //每个僵尸队伍的路线

mapmp; //给僵尸队伍坐标离散化为数字
int idx = 1;
int getidx(PII x){
    if(!mp.count(x))mp[x] = idx++;
    return mp[x];
}

int dist[maxn][maxn], vis[maxn][maxn], tim[maxn][maxn];//Dij
PII path[maxn][maxn];

int main(){
    ios::sync_with_stdio(0), cin.tie(0),cout.tie(0);
    //input&init
    cin>>n>>m>>T;
    for(int i = 0; i >a[i][j];
            if(a[i][j] n || jm && a[i][j]>0){
                js.push_back({i,j,a[i][j]});   //去掉四个点后的僵尸坐标和数量
            }
        }
    }
    //网格图上跑Dijkstra
    priority_queue, greater >q;
    q.push({0,{st.first,st.second}});   //消耗,坐标
    dist[st.first][st.second] = 0;
    while(q.size()){
        auto t = q.top();  q.pop();
        PII pos = t.second;
        if(vis[pos.first][pos.second])continue;
        vis[pos.first][pos.second] = 1;
        for(int i = 0; i  n || ny  m)continue;
            if(dist[nx][ny] > dist[pos.first][pos.second]+a[nx][ny]){//找消耗最小的
                dist[nx][ny] = dist[pos.first][pos.second]+a[nx][ny];
                path[nx][ny] = pos;
                tim[nx][ny] = tim[pos.first][pos.second]+1;
            }else if(dist[nx][ny] == dist[pos.first][pos.second]+a[nx][ny]){
                if(tim[nx][ny] > tim[pos.first][pos.second]+1){//消耗相等，找到大本营最快的
                    tim[nx][ny] = tim[pos.first][pos.second]+1;
                    path[nx][ny] = pos;
                }
            }
            q.push({dist[nx][ny], {nx,ny}});
        }
    }
    //枚举每个僵尸，更新最优路径线路
    for(auto it: js){
        int id = getidx({it.x,it.y});
        cnt[id] = it.cc;
        PII tmp;
        for(int i = 0; i =1 && nx=1 && nyse;  //维护正在被攻打的防御堡垒
        for(auto it : js){ //僵尸走路
            int id = getidx({it.x, it.y});//获取这支僵尸队伍的唯一编号
            if(cnt[id]==0)continue;
            int x = p[id][now[id]].first, y = p[id][now[id]].second;
            if(a[x][y]>0 && !(x==st.first && y==st.second)){
                se.insert({x,y});
            }else if(a[x][y]0)a[x][y]--;//攻打堡垒
                else if(a[x][y]