HDU - 3338 Kakuro Extension(最大流)

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題目大意：看一基本就知道了

解題思路：痛點是構圖。。
設定一個超級源點和所有的行的和相連，容量為該行的和 - 該行和由幾個數相加得到
設定一個超級匯點，和所有列的和相連，容量為該列的和 - 該列和由幾個數相加得到
接著就是空白部分和 “行和“和 “列和“的關係了
空白連向該行的行和，權值為8
空白連向該列的列和，權值也為8
為什麼為8，而不是9，因為流量也可能為0，但是0是不能填的，所以將容量設為8，最後取值的時候加1即可

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;#define N 100010#define INF 0x3f3f3f3fstruct Edge {    int from, to, cap, flow;    Edge() {}    Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}};struct ISAP {    int p[N], num[N], cur[N], d[N];    int t, s, n, m;    bool vis[N];    vector<int> G[N];    vector<Edge> edges;    void init(int n) {        this->n = n;        for (int i = 0; i <= n; i++) {            G[i].clear();            d[i] = INF;        }        edges.clear();    }    void AddEdge(int from, int to, int cap) {        edges.push_back(Edge(from, to, cap, 0));        edges.push_back(Edge(to, from, 0, 0));        m = edges.size();        G[from].push_back(m - 2);        G[to].push_back(m - 1);    }    bool BFS() {        memset(vis, 0, sizeof(vis));        queue<int> Q;        d[t] = 0;        vis[t] = 1;        Q.push(t);        while (!Q.empty()) {            int u = Q.front();            Q.pop();            for (int i = 0; i < G[u].size(); i++) {                Edge &e = edges[G[u][i] ^ 1];                if (!vis[e.from] && e.cap > e.flow) {                    vis[e.from] = true;                    d[e.from] = d[u] + 1;                    Q.push(e.from);                }            }        }        return vis[s];    }    int Augment() {        int u = t, flow = INF;        while (u != s) {            Edge &e = edges[p[u]];            flow = min(flow, e.cap - e.flow);            u = edges[p[u]].from;        }        u = t;        while (u != s) {            edges[p[u]].flow += flow;            edges[p[u] ^ 1].flow -= flow;            u = edges[p[u]].from;        }        return flow;    }    int Maxflow(int s, int t) {        this->s = s; this->t = t;        int flow = 0;        BFS();        if (d[s] > n)            return 0;        memset(num, 0, sizeof(num));        memset(cur, 0, sizeof(cur));        for (int i = 0; i < n; i++)            if (d[i] < INF)                num[d[i]]++;        int u = s;        while (d[s] <= n) {            if (u == t) {                flow += Augment();                u = s;            }            bool ok = false;            for (int i = cur[u]; i < G[u].size(); i++) {                Edge &e = edges[G[u][i]];                if (e.cap > e.flow && d[u] == d[e.to] + 1) {                    ok = true;                    p[e.to] = G[u][i];                     cur[u] = i;                    u = e.to;                    break;                }            }            if (!ok) {                int Min = n;                for (int i = 0; i < G[u].size(); i++) {                    Edge &e = edges[G[u][i]];                    if (e.cap > e.flow)                        Min = min(Min, d[e.to]);                }                if (--num[d[u]] == 0)                    break;                num[d[u] = Min + 1]++;                cur[u] = 0;                if (u != s)                    u = edges[p[u]].from;            }        }        return flow;    }};ISAP isap;#define M 110struct Node {    int row, col, val;    Node() {}    Node(int row, int col, int val): row(row), col(col), val(val){}}row[N], col[N];int n, m;int empty[M][M];char map[M];void solve() {    int cnt_row = 0, cnt_col = 0, cnt_empty = 0;    memset(empty, 0, sizeof(empty));    for (int i = 1; i <= n; i++)        for (int j = 1; j <= m; j++) {            scanf("%s", map);            if (strcmp(map, ".......") == 0)                empty[i][j] = ++cnt_empty;            else if(map[3] == ‘\\‘) {                int t;                if (map[0] != ‘X‘) {                    t = (map[0] - ‘0‘) * 100 + (map[1] - ‘0‘) * 10 + (map[2] - ‘0‘);                    col[++cnt_col] = Node(i, j, t);                }                if (map[4] != ‘X‘) {                    t = (map[4] - ‘0‘) * 100 + (map[5] - ‘0‘) * 10 + (map[6] - ‘0‘);                    row[++cnt_row] = Node(i, j, t);               }                }        }    int s = 0, t = cnt_empty + cnt_row + cnt_col + 1;    isap.init(t);    for (int i = 1; i <= cnt_row; i++) {        int x = row[i].row;        int cnt = 0;        for (int y = row[i].col + 1; y <= m; y++) {            if(empty[x][y]) {                cnt++;                isap.AddEdge(i, cnt_row + empty[x][y], 8);            }            else                break;        }        isap.AddEdge(s, i, row[i].val - cnt);    }    for (int i = 1; i <= cnt_col; i++) {        int y = col[i].col;        int cnt = 0;        for (int x = col[i].row + 1; x <= n; x++) {            if (empty[x][y]) {                cnt++;                isap.AddEdge(cnt_row + empty[x][y], cnt_row + cnt_empty + i, 8);            }            else                 break;        }        isap.AddEdge(cnt_row + cnt_empty + i, t, col[i].val - cnt);    }    isap.Maxflow(s, t);    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= m; j++) {            if (empty[i][j]) {                int ans = 0;                for (int k = 0; k < isap.G[empty[i][j] + cnt_row].size(); k++) {                    Edge &e = isap.edges[isap.G[empty[i][j] + cnt_row][k]];                    if (e.to > cnt_row + cnt_empty) {                        ans += e.flow;                        break;                    }                }                printf("%d ", ans + 1);            }            else                printf("_ ");        }        printf("\n");    }}int main() {    while (scanf("%d%d", &n, &m) != EOF) {        solve();    }    return 0;}

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HDU - 3338 Kakuro Extension(最大流)