1≤m,n≤20 1 \leq m, n \leq 201≤m,n≤20
根据题目描述建图:
第一问:
将每个点拆成出点和入点,入点和出点连边容量为1,保证每个点不重复,每个出点和左下角点以及右下角点连一容量为1
费用为目标点权值的边,保证了边不重复。源点与第一行所有点的入点连一条容量为1费用为目标点权值的边。为保证有且
只有m条路径,我的方法先让最后一行每个点与汇点1建边,容量为1费用为0,再让汇点1和最终汇点建一容量为m费用为0的边。
这样可以保证最终会产生m条路径。
第二问:
与第一问类似,但是允许重复利用同一节点,那么只要把每个节点的入点和出点之间的边的容量修改为INF就好了,其他同上。
第三问:
在第二问的基础上,把不同点之间所连边的容量改为INF即可。
三次询问分别求出最小费用最大流。
#include<stdio.h>#include<algorithm>#include<string.h>#include<queue>using namespace std;const int maxm = 10005;const int maxn = 100005;const int INF = 1e9 + 7;struct node{int u, v, flow, cost, next;}edge[maxn];int dis[maxm], head[maxm], cur[maxm], pre[maxn], f[1005][1005], map[1005][1005];int s, t, n, m, cnt;void init(){cnt = 0, s = 0, t = 10000;memset(head, -1, sizeof(head));}void add(int u, int v, int w, int cost){edge[cnt].u = u, edge[cnt].v = v;edge[cnt].flow = w, edge[cnt].cost = cost;edge[cnt].next = head[u], head[u] = cnt++;edge[cnt].u = v, edge[cnt].v = u;edge[cnt].flow = 0, edge[cnt].cost = -cost;edge[cnt].next = head[v], head[v] = cnt++;}int bfs(){queue<int>q;for (int i = 0;i <= 10004;i++) dis[i] = INF;memset(pre, -1, sizeof(pre));dis[s] = 0;q.push(s);int rev = 0;while (!q.empty()){int u = q.front();q.pop();for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].v;if (dis[v] > dis[u] + edge[i].cost&&edge[i].flow){dis[v] = dis[u] + edge[i].cost;pre[v] = i;q.push(v);}}}if (dis[t] == INF) return 0;return 1;}int MCMF(){int minflow, ans = 0;while (bfs()){minflow = INF;for (int i = pre[t];i != -1;i = pre[edge[i].u])minflow = min(minflow, edge[i].flow);for (int i = pre[t];i != -1;i = pre[edge[i].u]){edge[i].flow -= minflow;edge[i ^ 1].flow += minflow;}ans += minflow*dis[t];}return ans;}int main(){int i, j, k, sum, id = 0;scanf("%d%d", &m, &n);init();for (i = 1;i <= n;i++){for (j = 1;j <= m + i - 1;j++){scanf("%d", &f[i][j]);map[i][j] = ++id;}}for (i = 1;i <= n;i++){for (j = 1;j <= m + i - 1;j++){if (i == 1) add(s, map[i][j], 1, -f[i][j]);if (i == n) add(map[i][j] + id, id * 2 + 1, 1, 0);add(map[i][j], map[i][j] + id, 1, 0);if (i < n){add(map[i][j] + id, map[i + 1][j], 1, -f[i + 1][j]);add(map[i][j] + id, map[i + 1][j + 1], 1, -f[i + 1][j + 1]);}}}add(id * 2 + 1, t, m, 0);printf("%d\n", -MCMF());init();for (i = 1;i <= n;i++){for (j = 1;j <= m + i - 1;j++){if (i == 1) add(s, map[i][j], 1, -f[i][j]);if (i == n) add(map[i][j] + id, id * 2 + 1, INF, 0);add(map[i][j], map[i][j] + id, INF, 0);if (i < n){add(map[i][j] + id, map[i + 1][j], 1, -f[i + 1][j]);add(map[i][j] + id, map[i + 1][j + 1], 1, -f[i + 1][j + 1]);}}}add(id * 2 + 1, t, m, 0);printf("%d\n", -MCMF());init();for (i = 1;i <= n;i++){for (j = 1;j <= m + i - 1;j++){if (i == 1) add(s, map[i][j], 1, -f[i][j]);if (i == n) add(map[i][j] + id, id * 2 + 1, INF, 0);add(map[i][j], map[i][j] + id, INF, 0);if (i < n){add(map[i][j] + id, map[i + 1][j], INF, -f[i + 1][j]);add(map[i][j] + id, map[i + 1][j + 1], INF, -f[i + 1][j + 1]);}}}add(id * 2 + 1, t, m, 0);printf("%d\n", -MCMF());return 0;}