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题意】有n(<=20)只队伍比赛, 队伍i初始得分w[i], 剩余比赛场数r[i](包括与这n只队伍以外的队伍比赛), remain[i][j]表示队伍i与队伍j剩余比赛场数, 没有平局, 问队伍1有没有可能获得这n队中的第一名(可以有并列第一). 【
构图方法】 源点向队伍节点连流量为X的边表示该队伍最多赢X场；两队间比赛节点向汇点连流量为Y的边表示这两队间要进行Y场比赛，两队伍节点向对应比赛节点各连一条流量为Z的边表示每个队最多赢对方Z场 【
思路】 按照上面的思路建图求出最大流，如果是满流则表示比赛可以安排，便为YES. 注意： 一、队伍同其他分区队伍的比赛可以不用管，认为他们全都输掉就可以了. 二、第一个队伍让他全赢就可以了，网络流中只需要建其他N-1个队伍的节点，比赛也是N-1个队伍之间的比赛，不需要管第一支队伍了.
#include #include #include #include #include #include #define MID(x,y) ((x+y)/2)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const int MAXV = 505;const int MAXE = 10005;const int oo = 0x3fffffff;struct node{ int u, v, flow; int opp; int next;};struct Dinic{ node arc[MAXE]; int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数 int cur[MAXV]; //当前弧 int q[MAXV]; //bfs建层次图时的队列 int path[MAXE], top; //存dfs当前最短路径的栈 int dep[MAXV]; //各节点层次 void init(int n){ vn = n; en = 0; mem(head, -1); } void insert_flow(int u, int v, int flow){ arc[en].u = u; arc[en].v = v; arc[en].flow = flow; arc[en].opp = en + 1; arc[en].next = head[u]; head[u] = en ++; arc[en].u = v; arc[en].v = u; arc[en].flow = 0; //反向弧 arc[en].opp = en - 1; arc[en].next = head[v]; head[v] = en ++; } bool bfs(int s, int t){ mem(dep, -1); int lq = 0, rq = 1; dep[s] = 0; q[lq] = s; while(lq < rq){ int u = q[lq ++]; if (u == t){ return true; } for (int i = head[u]; i != -1; i = arc[i].next){ int v = arc[i].v; if (dep[v] == -1 && arc[i].flow > 0){ dep[v] = dep[u] + 1; q[rq ++] = v; } } } return false; } int solve(int s, int t){ int maxflow = 0; while(bfs(s, t)){ int i, j; for (i = 1; i <= vn; i ++) cur[i] = head[i]; for (i = s, top = 0;;){ if (i == t){ int mink; int minflow = 0x3fffffff; for (int k = 0; k < top; k ++) if (minflow > arc[path[k]].flow){ minflow = arc[path[k]].flow; mink = k; } for (int k = 0; k < top; k ++) arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow; maxflow += minflow; top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内). i = arc[path[top]].u; } for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){ int v = arc[j].v; if (arc[j].flow && dep[v] == dep[i] + 1) break; } if (j != -1){ path[top ++] = j; i = arc[j].v; } else{ if (top == 0) break; dep[i] = -1; i = arc[path[-- top]].u; } } } return maxflow; }}dinic;int win[25];int remain[25][25];int contest;int con[MAXV][MAXV]; //记录i和j的比赛节点号int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); int n; scanf("%d", &n); int max_win = 0; for (int i = 1; i <= n; i ++){ scanf("%d", &win[i]); max_win = max(max_win, win[i]); } for (int i = 1; i <= n; i ++){ int tmp; scanf("%d", &tmp); if (i == 1){ win[1] += tmp; } } if (win[1] < max_win){ puts("NO"); return 0; } contest = 1; for (int i = 1; i <= n; i ++){ for (int j = 1; j <= n; j ++){ if (i > j){ con[i][j] = con[j][i] = contest ++; } scanf("%d", &remain[i][j]); } } int node_num = n + n*(n-1)/2; int sum = 0; dinic.init(node_num+2); for (int i = 2; i <= n; i ++){ dinic.insert_flow(node_num+1, i, win[1] - win[i]); //源点向N-1个队伍连一条边表示每支队伍最多能赢几场 } for (int i = 2; i <= n; i ++){ for (int j = i+1; j <= n; j ++){ dinic.insert_flow(n+con[i][j], node_num+2, remain[i][j]); //两两队伍间的比赛建一个节点，向汇点连一条边表示两个队伍将进行几场比赛 dinic.insert_flow(i, n+con[i][j], remain[i][j]); //i向该比赛连一条边表示i能赢几场 dinic.insert_flow(j, n+con[i][j], remain[i][j]); //j向该比赛连一条边表示i能赢几场 sum += remain[i][j]; } } if (dinic.solve(node_num+1, node_num+2) == sum){ puts("YES"); } else{ puts("NO"); } return 0;}