计蒜客2015 初赛题解及代码

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2015 计蒜客初赛题解之工作区的颜值选择

标签（空格分隔）： ACM

—题解思路

简单数据暴力代码

#include <iostream>#include <cstdio>using namespace std;const int Maxn = 2, Maxm = 3;int C[Maxn][Maxm], U[Maxn][Maxm], W[Maxn][Maxm];int n, m, k;inline int abs(int x) {    return x < 0 ? -x : x;}inline int cost(int c, int i, int j) {    return (abs(c + i + j + 2) ^ U[i][j]) * W[i][j];}int dirx[] = {-1, 0, 1, 0};int diry[] = {0, -1, 0, 1};int optimize(int x, int y) {    if (y >= m) return 0;    if (x >= n) return optimize(x ^ 1, y + 1);    int best = -1;    for (int i = -k; i <= k; ++i) {        int tmp = cost(i,x, y);        for (int l = 0; l < 4; ++l) {            int nx = dirx[l] + x;            int ny = diry[l] + y;            if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;            tmp += abs(i + C[nx][ny]);        }        if (best == -1 || tmp < best) best = tmp;    }    return best + optimize(x ^ 1, y + 1);}int solve(int x, int y) {    if (y >= m) return optimize(1, 0);    if (x >= n) return solve(x ^ 1, y + 1);    int ans = -1;    for (int i = -k; i <= k; ++i) {        C[x][y] = i;        if (ans == -1) ans = solve(x ^ 1, y + 1) + cost(i, x, y);        else ans = min(ans, solve(x ^ 1, y + 1) + cost(i, x, y));    }    return ans;}int main(){    scanf("%d %d %d", &n, &m, &k);    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d", &U[i][j]);        }    }    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d", &W[i][j]);        }    }    printf("%d\n", solve(0, 0));    return 0;}

中等数据动态规划代码（题目范围出错 n可以等于1）

#include <iostream>#include <cstdio>using namespace std;const int Maxn = 2, Maxm = 33, Maxk=33, INF = 0x7FFFFFFF;int U[Maxn][Maxm], W[Maxn][Maxm];int n, m, k;inline int abs(int x) {    return x < 0 ? -x : x;}inline int cost(int c, int i, int j) {    return (abs(c + i + j + 2) ^ U[i][j]) * W[i][j];}inline int ID(int x) {    return x + Maxk;}int dp[Maxm][Maxk << 1][Maxk << 1];int neg_decision[Maxm][Maxk << 1][Maxk << 1];int pos_decision[Maxm][Maxk << 1][Maxk << 1];int solve() {    for (int i = -k; i <= k; ++i) {        for (int j = -k; j <= k; ++j) {            dp[0][ID(i)][ID(j)] = cost(i, 0, 0) + cost(j, 1, 0) + abs(i + j);        }        int neg = -k;        int pos = k;        for (int j = -k; j <= k; ++j) {            if (dp[0][ID(i)][ID(j)] - j < dp[0][ID(i)][ID(neg)] - neg)                neg_decision[0][ID(i)][ID(j)] = j;            else neg_decision[0][ID(i)][ID(j)] = neg;            if (dp[0][ID(i)][ID(-j)] - j < dp[0][ID(i)][ID(pos)] + pos)                pos_decision[0][ID(i)][ID(-j)] = -j;            else pos_decision[0][ID(i)][ID(-j)] = pos;            neg = neg_decision[0][ID(i)][ID(j)];            pos = pos_decision[0][ID(i)][ID(-j)];        }    }    for (int t = 1; t < m; ++t) {        for (int i = -k; i <= k; ++i) {            for (int j = -k; j <= k; ++j){                int& tmp = dp[t][ID(i)][ID(j)];                 tmp = INF;                for (int l = -k; l <= k; ++l) {                    int _neg = neg_decision[t - 1][ID(l)][ID(-j)];                    int _pos = pos_decision[t - 1][ID(l)][ID(-j)];                    tmp = min(dp[t - 1][ID(l)][ID(_neg)] - _neg - j  + abs(l + i), tmp);                    tmp = min(dp[t - 1][ID(l)][ID(_pos)] + _pos + j  + abs(l + i), tmp);                }                tmp += abs(i + j) + cost(i, 0, t) + cost(j, 1, t);            }            int neg = -k;            int pos = k;            for (int j = -k; j <= k; ++j){                if (dp[t][ID(i)][ID(j)] - j < dp[t][ID(i)][ID(neg)] - neg)                    neg_decision[t][ID(i)][ID(j)] = j;                else neg_decision[t][ID(i)][ID(j)] = neg;                if (dp[t][ID(i)][ID(-j)] - j < dp[t][ID(i)][ID(pos)] + pos)                    pos_decision[t][ID(i)][ID(-j)] = -j;                else pos_decision[t][ID(i)][ID(-j)] = pos;                neg = neg_decision[t][ID(i)][ID(j)];                pos = pos_decision[t][ID(i)][ID(-j)];            }        }    }    int ans = INF;    for (int i = -k; i <= k; ++i)        for (int j = -k; j <= k; ++j)            ans = min(ans, dp[m - 1][ID(i)][ID(j)]);    return ans;}int dp1[Maxm][Maxk << 1];int solve_n_1() {    for (int i = -k; i <= k; ++i)        dp1[0][ID(i)] = cost(i, 0, 0);    for (int t = 1; t < m; ++t) {        for (int i = -k; i <= k; ++i) {            dp1[t][ID(i)] = INF;            for (int j = -k; j <= k; ++j) {                dp1[t][ID(i)] = min(dp1[t][ID(i)], dp1[t - 1][ID(j)] + abs(i + j) + cost(i, 0, t));            }        }    }    int ans = INF;    for (int i = -k; i <= k; ++i)        ans = min(ans, dp1[m - 1][ID(i)]);    return ans;}int main(){    scanf("%d %d %d", &n, &m, &k);    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d", &U[i][j]);        }    }    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d", &W[i][j]);        }    }    if (n == 1) printf("%d\n", solve_n_1());    else printf("%d\n", solve());    return 0;}

困难数据网络流代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 100000, INF = (1 << 30) - 1;const int Maxn = 33, Maxm = 33, Maxk=33;int U[Maxn][Maxm], W[Maxn][Maxm];int n, m, k;inline int abs(int x) {    return x < 0 ? -x : x;}inline int cost(int c, int i, int j) {    return (abs(c + i + j + 2) ^ U[i][j]) * W[i][j];}struct Edge {    int to, f, next;    Edge(){}    Edge(int to, int f, int next): to(to), f(f), next(next){}}edge[MAXN * 4];int box[MAXN], size;void add(int from, int to, int f){    edge[size] = Edge(to, f, box[from]);    box[from] = size++;}int deep[MAXN],depnum[MAXN];// depnum记录各个距离的个数bool isEnd;int s,t,N;void bfs()//反向分层{    memset(deep,-1,sizeof(deep));    memset(depnum,0,sizeof(depnum));    deep[t]=0;    depnum[0]++;    int que[MAXN],in=0,out=0;    que[in++]=t;    while(in>out)    {        int x=que[out++];        for(int i=box[x];i!=-1;i=edge[i].next)        {            int x1=edge[i].to;            if(edge[i^1].f==0||deep[x1]!=-1)continue;            que[in++]=x1;            deep[x1]=deep[x]+1;            depnum[deep[x1]]++;        }    }    if(deep[s]==-1) isEnd=true;    else isEnd=false;}int dfs(int x,int flow){    if(x==t)return flow;     if(isEnd||flow==0)return 0;    int tmp=0;    int min1=deep[x]+1;    for(int i=box[x];i!=-1;i=edge[i].next)    {        int x1=edge[i].to;      if(edge[i].f==0||deep[x1]==N||deep[x1]==-1)continue;        if(deep[x1]==deep[x]-1)//如果有有效路径则更新下一结点        {            int tmp1=dfs(x1,min(edge[i].f,flow-tmp));            edge[i].f-=tmp1;            edge[i^1].f+=tmp1;        tmp+=tmp1;          }        if(edge[i].f)min1=min(min1,deep[x1]+1);    }      depnum[min1]++;    depnum[deep[x]]--;    if(depnum[deep[x]]==0)isEnd=true;    deep[x]=min1;    return tmp;}int sap(){    //N=n+2;    //s=n,t=n+1;    int ans=0;    bfs();    while(deep[s]<N&&!isEnd)    {        ans+=dfs(s,INF);    }    return ans;}int solve() {    memset(box, -1, sizeof(box)); size = 0;    s = 0;    t = (2 * k + 2) * n * m + 1;    int id = 0;    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            int first = id * (2 * k + 2) + 1;            int last = first + 2 * k + 1;            for (int l = first; l < last; ++l) {                add(l, l + 1, cost(-k + l - first, i, j));                add(l + 1, l, cost(-k + l - first, i, j));            }            if ((i + j) & 1) swap(first, last);            add(s, first, INF);            add(first, s, 0);            add(last, t, INF);            add(t, last, 0);              id++;        }    }    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            int cur = (i * m + j) * (2 * k + 2) + 1;            if (j) {                int left = (i * m + j - 1) * (2 * k + 2) + 1;                for (int l = 0; l < 2 * k + 2; ++l) {                    add (left + 2 * k + 1 - l, cur + l, 1);                    add (cur + l, left + 2 * k + 1 - l, 1);                }            }            if (i) {                int up = ((i - 1) * m + j) * (2 * k + 2) + 1;                for (int l = 0; l < 2 * k + 2; ++l) {                    add (up + 2 * k + 1 - l, cur + l, 1);                    add (cur + l, up + 2 * k + 1 - l, 1);                }            }        }    }    N = t + 1;    return sap();}int main(){    scanf("%d %d %d", &n, &m, &k);    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d", &U[i][j]);        }    }    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            scanf("%d", &W[i][j]);        }    }    printf("%d\n", solve());    return 0;}

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