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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).
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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
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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;
}