hdu 4971

来源：互联网发布：下载贵金属软件编辑：程序博客网时间：2024/06/06 17:43
经典闭权和问题，正权的点由源点向该点连边，负权的点由该点向负权连边，做最大流，找到最小割即可，然后由正权点权和减去最小割容量。
AC代码：
#include <cstdio>#include <cstring>#include <map>#include <vector>#include <algorithm>using namespace std;const int maxn = 10007;const int INF = 10000007;int n, m;int head[maxn], edge_number, cur[maxn], dist[maxn], s[maxn], que[maxn], source, dest, flow;struct Edge {    int u, v, cap, next;};struct Edge edge[maxn];void addEdge(int u, int v, int cap) {    edge[edge_number].u = u;    edge[edge_number].v = v;    edge[edge_number].cap = cap;    edge[edge_number].next = head[u];    head[u] = edge_number;    edge_number++;    edge[edge_number].u = v;    edge[edge_number].v = u;    edge[edge_number].cap = 0;    edge[edge_number].next = head[v];    head[v] = edge_number;    edge_number++;}int bfs() {    int front = 0;    int rear = 1;    memset(dist, -1, sizeof(dist));    dist[source] = 0;    que[front] = source;    while (front < rear) {        int cnt = que[front];        for (int i = head[cnt]; i != -1; i = edge[i].next) {            int v = edge[i].v;            if (edge[i].cap > 0 && dist[v] == -1) {                dist[v] = dist[cnt] + 1;                que[rear++] = v;                if (v == dest) {                    return 1;                }            }        }        front++;    }    return 0;}int dinic() {    int top, cnt, min_flow, min_flow_num, flow;    flow = 0;    while (bfs()) {        memcpy(cur, head, sizeof(cur));        top = 0;        cnt = source;        while (1) {            if (cnt == dest) {                min_flow = INF;                for (int i = 0; i < top; i++) {                    if (edge[s[i]].cap < min_flow) {                        min_flow = edge[s[i]].cap;                        min_flow_num = i;                    }                }                for (int i = 0; i < top; i++) {                    edge[s[i]].cap -= min_flow;                    edge[s[i] ^ 1].cap += min_flow;                }                flow += min_flow;                top = min_flow_num;                cnt = edge[s[top]].u;            }            for (int i = cur[cnt]; i != -1; cur[cnt] = i = edge[i].next) {                if (edge[i].cap > 0 && dist[edge[i].v] == dist[cnt] + 1) {                    break;                }            }            if (cur[cnt] != -1) {                s[top++] = cur[cnt];                cnt = edge[cur[cnt]].v;            }            else {                if (top == 0) {                    break;                }                dist[cnt] = -1;                top--;                cnt = edge[s[top]].u;            }        }    }    return flow;}int main() {//    freopen("in.txt", "r", stdin);    int test_case;    scanf("%d", &test_case);    int ncase = 1;    while (test_case--) {        edge_number = 0;        memset(head, -1, sizeof(head));        scanf("%d%d", &n, &m);        int sums = 0;        source = 0;        dest = n + m + 1;        for (int i = 1; i <= n; i++) {            int x;            scanf("%d", &x);            addEdge(source, i, x);            sums += x;        }        for (int i = 1; i <= m; i++) {            int x;            scanf("%d", &x);            addEdge(n + i, dest, x);        }        for (int i = 1; i <= n; i++) {            int a;            scanf("%d", &a);            for (int j = 1; j <= a; j++) {                int c;                scanf("%d", &c);                addEdge(i, n + 1 + c, INF);            }        }        for (int i = 0; i < m; i++) {            for (int j = 0; j < m; j++) {                int x;                scanf("%d", &x);                if (x == 1) {                    addEdge(n + 1 + i, n + 1 + j, INF);                }            }        }        int ans = dinic();        printf("Case #%d: %d\n", ncase, sums - ans);        ncase++;    }    return 0;}
0 0