HDU 4888 Redraw Beautiful Drawings（2014 Multi-University Training Contest 3）

题意：给定n*m个格子，每个格子能填0-k 的整数。然后给出每列之和和每行之和，问有没有解，有的话是不是唯一解，是唯一解输出方案。

思路：网络流，一共 n+m+2个点   源点 到行连流量为 所给的 当前行之和。    每行 连到每一列 一条流量为  k的边，每列到汇点连 列和。如果流量等于总和则有解，反之无解（如果列总和不等于行总和也无解）。  判断方案是否唯一 找残留网络是否存在长度大于2的环即可，有环说明不唯一。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include <iostream>
#include<climits>
using namespace std;
const int N = 1000;
const int M = 1000000;
int n;
int ecnt, head[N], nx[M], to[M], va[M], cur_edge[N];
int source, target, flow, pre[N], lev[N], qu[N], sign;
void addedge(int u, int v, int w) {
    to[ecnt] = v;
    nx[ecnt] = head[u];
    va[ecnt] = w;
    head[u] = ecnt++;
}
bool bfs(int s, int t) {
    std::fill(lev, lev + n, -1);
    sign = t;
    lev[t] = 0;
    int st = 0, ed = 0;
    qu[ed++] = t;
    while (st != ed && lev[s] == -1) {
        int u = qu[st++];
        for (int i = head[u]; i != -1; i = nx[i]) {
            if (va[i ^ 1] > 0 && lev[to[i]] == -1) {
                lev[to[i]] = lev[u] + 1;
                qu[ed++] = to[i];
            }
        }
    }
    return lev[s] != -1;
}
void push() {
    int delta = INT_MAX, u, p;
    for (u = target; u != source; u = to[p ^ 1]) {
        p = pre[u];
        delta = std::min(delta, va[p]);
    }
    for (u = target; u != source; u = to[p ^ 1]) {
        p = pre[u];
        va[p] -= delta;
        if (!va[p]) {//注意double时要改
            sign = to[p ^ 1];
        }
        va[p ^ 1] += delta;
    }
    flow += delta;
}
void dfs(int u) {
    if (u == target)
        push();
    else {
        for (int i = cur_edge[u]; i != -1; i = nx[i]) {
            if (va[i] > 0 && lev[u] == lev[to[i]] + 1) {
                pre[to[i]] = i;
                dfs(to[i]);
                if (lev[sign] > lev[u]) {
                    return;
                }
                sign = target;
            }
        }
        lev[u] = -1;
    }
}
void dinic(int s, int t, int node_cnt) {
    source = s;
    target = t;
    n = node_cnt;

    //construct graph

    flow = 0;
    while (bfs(source, target)) {
        for (int i = 0; i < n; ++i) {
            cur_edge[i] = head[i];
        }
        dfs(source);
    }

}
int nc, kc, mc, sum1, sum2;
int r[410], c[410];
void init() {
    sum1 = sum2 = 0;
    for (int i = 0; i < nc; ++i) {
        scanf("%d", &r[i]);
        sum1 += r[i];
    }
    for (int i = 0; i < mc; ++i) {
        scanf("%d", &c[i]);
        sum2 += c[i];
    }
}
int flag = 0;
bool bo[1000];
bool bb[1000];
int cas[350000];
int ri = 0;
void gao(int now, int fa) {//找环
    if(flag)return;
    //printf("->%d ",now);
    bo[now] = true;
    for (int i = head[now]; i != -1; i = nx[i]) {
        int u = to[i];
        if (u == fa)
            continue;
        if (va[i] == 0)
            continue;

        //    printf(" {%d %d}\n",now,u);
        if (bb[u]) {
            //    puts("fuck");
            flag = 1;
            return;
        }
        if (cas[i] == ri)
            continue;
        cas[i] = ri;
        //if(bo[u])continue;
        bb[u] = true;
        gao(u, now);
        bb[u] = false;
    }
}
void solve() {
    if (sum1 != sum2) {
        puts("Impossible");
        return;
    }
    std::fill(head, head + mc + nc + 5, -1);
    ecnt = 0;
    int s, t;
    s = nc + mc;
    t = s + 1;
    //    printf("st:%d %d\n",s,t);
    for (int i = 0; i < nc; ++i) {
        for (int j = 0; j < mc; ++j) {

            //        printf("[%d %d]\n",i,j+nc);
            addedge(i, j + nc, kc);
            addedge(j + nc, i, 0);
        }
    }
    for (int i = 0; i < nc; ++i) {
        //    printf("[%d %d]\n",s,i);
        addedge(s, i, r[i]);
        addedge(i, s, 0);
    }
    for (int i = 0; i < mc; ++i) {

        //    printf("[%d %d]\n",i+nc,t);
        addedge(i + nc, t, c[i]);
        addedge(t, i + nc, 0);
    }
    dinic(s, t, t + 2);
    //    printf("flow:%d\n",flow);
    if (flow == sum1) {

        memset(bo, 0, sizeof(bo));
        memset(bb, 0, sizeof(bb));
        flag = 0;
        for (int i = 0; i <= t; ++i) {
            if(flag)break;
            bb[i] = true;
            gao(i, -1);
            bb[i] = false;
        }
        if (flag)
            puts("Not Unique");
        else {
            puts("Unique");
            for (int i = 0; i < nc; ++i) {
                for (int j = 0; j < mc; ++j) {
                    int now = i * mc + j;
                    printf("%d", va[now << 1 | 1]);
                    if (j == mc - 1)
                        puts("");
                    else
                        printf(" ");
                }
            }
        }
    } else
        puts("Impossible");
}
int main() {
    memset(cas, 0, sizeof(cas));
    while (scanf("%d%d%d", &nc, &mc, &kc) != EOF) {
        ++ri;
        init();
        solve();
    }
    return 0;
}

HDU 4888 Redraw Beautiful Drawings（2014 Multi-University Training Contest 3）,布布扣,bubuko.com