[poj3159]

题意：n个人，m个信息，每行的信息是3个数字，A,B,C，表示B比A多出来的糖果不超过C个，问你，n号人最多比1号人多几个糖果

solution

spfa+差分约束系统

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <stack>using namespace std;#define N 30010#define M 150010int nxt[M], to[M], vl[M], head[N], dist[N];int n, cnt = 0;bool vis[N];void addeage( int u, int v, int w ){    nxt[++cnt] = head[u], to[cnt] = v, vl[cnt] = w, head[u] = cnt;}int spfa( int t ){    stack<int> q;    while ( !q.empty() ) q.pop();    q.push(1);    memset( dist, 0x3f3f3f, sizeof(dist));    memset( vis, false, sizeof(vis));    dist[1] = 0;    vis[1] = true;    while ( !q.empty() ){        int cur = q.top();        q.pop();        vis[ cur ] = false;        for ( int i = head[ cur ]; i; i = nxt[i]){            int v = to[i], w = vl[i];            if ( dist[v] > dist[cur] + w ){                dist[v] = dist[cur] + w;                if ( !vis[v] ){                    vis[v] = true;                    q.push( v );                }            }        }    }    return dist[t];}int main(){    int n, m, a, b, c;    scanf( "%d%d", &n, &m);    cnt = 0;    for ( int i = 1; i <= m; i++ ){        scanf( "%d%d%d", &a, &      b, &c);        addeage( a, b, c);    }    printf( "%d", spfa(n));    return 0;}